Strategi A+ Ting 5 - Matematik Tambahan

. BHD. PENERBITDN. BHD. PENERBITSDN. BHD. PENERBDN. BHD. PENE

1 Rekod Prestasi Murid R1 Bab 1 Sukatan Membulat Revisi Pantas 1 Praktis PBD 1.1 Radian 2 1.2 Panjang Lengkok Suatu Bulatan 3 1.3 Luas Sektor Suatu Bulatan 8 1.4 Aplikasi Sukatan Membulat 13 Praktis Berformat SPM 14 Zon KBAT 16 Bab 2 Pembezaan Revisi Pantas 17 Praktis PBD 2.1 Had dan Hubungannya dengan Pembezaan 19 2.2 Pembezaan Peringkat Pertama 23 2.3 Pembezaan Peringkat Kedua 29 2.4 Aplikasi Pembezaan 31 Praktis Berformat SPM 47 Zon KBAT 49 Bab 3 Pengamiran Revisi Pantas 50 Praktis PBD 3.1 Pengamiran sebagai Songsangan Pembezaan 52 3.2 Kamiran Tak Tentu 53 3.3 Kamiran Tentu 58 3.4 Aplikasi Pengamiran 73 Praktis Berformat SPM 74 Zon KBAT 77 Bab 4 Pilih Atur dan Gabungan Revisi Pantas 78 Praktis PBD 4.1 Pilih Atur 79 4.2 Gabungan 86 Praktis Berformat SPM 89 Zon KBAT 90 Bab 5 Taburan Kebarangkalian Revisi Pantas 91 Praktis PBD 5.1 Pemboleh Ubah Rawak 92 5.2 Taburan Binomial 98 5.3 Taburan Normal 105 Praktis Berformat SPM 110 Zon KBAT 112 Bab 6 Fungsi Trigonometri Revisi Pantas 113 Praktis PBD 6.1 Sudut Positif dan Sudut Negatif 115 6.2 Nisbah Trigonometri bagi Sebarang Sudut 116 6.3 Graf Fungsi Sinus, Kosinus dan Tangen 118 6.4 Identiti Asas 122 6.5 Rumus Sudut Majmuk dan Rumus Sudut Berganda 124 6.6 Aplikasi Fungsi Trigonometri 133 Praktis Berformat SPM 137 Zon KBAT 139 Bab 7 Pengaturcaraan Linear Revisi Pantas 140 Praktis PBD 7.1 Model Pengaturcaraan Linear 141 7.2 Aplikasi Pengaturcaraan Linear 150 Praktis Berformat SPM 153 Zon KBAT 155 Bab 8 Kinematik Gerakan Linear Revisi Pantas 156 Praktis PBD 8.1 Sesaran, Halaju dan Pecutan sebagai Fungsi Masa 157 8.2 Pembezaan dalam Kinematik Gerakan Linear 160 8.3 Pengamiran dalam Kinematik Gerakan Linear 164 8.4 Aplikasi Kinematik Gerakan Linear 169 Praktis Berformat SPM 172 Zon KBAT 173 Kandungan Jawapan 174 Strategi A+ Maths Tam Tg5-Kan_vim_3p.indd Cyan.indd 1 7/11/2023 9:47:15 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

1 1.1 Radian Radian 1 Sudut yang tercangkum di pusat sebuah bulatan ialah 1 radian jika panjang lengkok adalah sama dengan jejari bulatan itu. The angle subtended at the centre of a circle is 1 radian if the arc length is equal to the radius of the circle. j unit j unit 1 rad B A O 2 Hubungan antara ukuran sudut radian dan darjah ialah The relationship between radians and degrees is 2π rad = 360° π rad = 180° 3 Rumus untuk menukarkan radian kepada darjah dan sebaliknya: The formula to convert radians to degrees and vice versa: a rad = (a × 180° π ) darjah/ degrees a° = (a × π 180 ) rad 1.2 Panjang Lengkok Suatu Bulatan Arc Length of a Circle 1 Jika s ialah panjang lengkok bagi sebuah bulatan yang berjejari j dan mencangkum sudut θ radian pada pusat O, maka If s is the arc length of a circle with the radius, j and the angle subtended by the arc at the centre is θ radian, then s = jθ Q s P j O θ 2 Tembereng merupakan rantau yang dibatasi oleh perentas dan panjang lengkok. Segment is the region bounded by the chord and the arc length. Q P R j O θ Perentas Chord Tembereng Segment Perimeter tembereng berlorek PQR Perimeter of the shaded segment PQR = Panjang perentas PR + Panjang lengkok PQR Length of chord PR + Arc length PQR = 2j 2 – 2j 2 kos/cos θ + jθ θ dalam darjah θ in degrees θ dalam radian θ in radian Kaedah Alternatif Perimeter tembereng berlorek PQR Perimeter of the shaded segment PQR = Panjang perentas PR + Panjang lengkok PQR Length of chord PR + Arc length PQR = 2j sin θ 2 + jθ θ dalam darjah θ in degrees θ dalam radian θ in radian 1.3 Luas Sektor Suatu Bulatan Area of Sector of a Circle 1 Dalam sebuah bulatan, sektor yang lebih kecil dikenali sebagai sektor minor, manakala sektor yang lebih besar dikenali sebagai sektor major. In a circle, the smaller sector is known as the minor sector, while the larger sector is known as the major sector. θ 2π – θ j Sektor minor Minor sector Sektor major Major sector Luas sektor minor = 1 2 j2 θ Area of minor sector Luas sektor major = 1 2 j2 (2π – θ) Area of major sector 2 Luas tembereng berlorek PTQ dapat dicari dengan menolak luas segi tiga OPQ daripada luas sektor POQ. The area of segment PTQ can be determined by subtracting the area of triangle OPQ from the area of sector POQ. Q P j T O θ Luas tembereng berlorek PTQ Area of the shaded segment PTQ = Luas sektor POQ – Luas segi tiga OPQ Area of sector POQ – Area of triangle OPQ = 1 2 j 2 θ – 1 2 j 2 sin θ = 1 2 j 2 (θ – sin θ) θ dalam radian θ in radian θ dalam darjah θ in degrees Sukatan Membulat Circular Measure Bidang Pembelajaran: Geometri Bab 1 1.1 Radian Radian 1 Sudut yang tercangkum di pusat sebuah bulatan ialah 1 radian jika panjang lengkok adalah sama dengan jejari bulatan itu. The angle subtended at the centre of a circle is 1 radian if the arc length is equal to the radius of the circle. j unit j unit 1 rad B A O 2 Hubungan antara ukuran sudut radian dan darjah ialah The relationship between radians and degrees is 2π rad = 360° π rad = 180° 3 Rumus untuk menukarkan radian kepada darjah dan sebaliknya: The formula to convert radians to degrees and vice versa: a rad = (a × 180° π ) darjah/ degrees a° = (a × π 180 ) rad 1.2 Panjang Lengkok Suatu Bulatan Arc Length of a Circle 1 Jika s ialah panjang lengkok bagi sebuah bulatan yang berjejari j dan mencangkum sudut θ radian pada pusat O, maka If s is the arc length of a circle with the radius, j and the angle subtended by the arc at the centre is θ radian, then s = jθ Q s P j O θ Q 2 Tembereng merupakan rantau yang dibatasi oleh perentas dan panjang lengkok. Segment is the region bounded by the chord and the arc length. Q P R j O θ Perentas Chord Tembereng Segment Perimeter tembereng berlorek PQR Perimeter of the shaded segment PQR = Panjang perentas PR + Panjang lengkok PQR Length of chord PR + Arc length PQR = 2j 2 – 2j 2 kos/cos θ + jθ θ dalam darjah θ in degrees θ dalam radian θ in radian Kaedah Alternatif Perimeter tembereng berlorek PQR Perimeter of the shaded segment PQR = Panjang perentas PR + Panjang lengkok PQR Length of chord PR + Arc length PQR = 2j sin θ 2 + jθ θ dalam darjah θ in degrees θ dalam radian θ in radian 1.3 Luas Sektor Suatu Bulatan Area of Sector of a Circle 1 Dalam sebuah bulatan, sektor yang lebih kecil dikenali sebagai sektor minor, manakala sektor yang lebih besar dikenali sebagai sektor major. In a circle, the smaller sector is known as the minor sector, while the larger sector is known as the major sector. θ 2π – θ j Sektor minor Minor sector Sektor major Major sector Luas sektor minor = 1 2 j2 θ Area of minor sector Luas sektor major = 1 2 j2 (2π – θ) Area of major sector 2 Luas tembereng berlorek PTQ dapat dicari dengan menolak luas segi tiga OPQ daripada luas sektor POQ. The area of segment PTQ can be determined by subtracting the area of triangle OPQ from the area of sector POQ. Q P j T O θ Luas tembereng berlorek PTQ Area of the shaded segment PTQ = Luas sektor POQ – Luas segi tiga OPQ Area of sector POQ – Area of triangle OPQ = 1 2 j 2 θ – 1 2 j 2 sin θ = 1 2 j 2 (θ – sin θ) θ dalam radian θ in radian θ dalam darjah θ in degrees Revisi Pantas Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 1 30/10/2023 8:45:20 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

2 Praktis PBD 1.1 Radian/ Radian Latihan 1 Tukar setiap sudut yang berikut kepada darjah. [Guna π = 3.142] TP 2 Convert each of the following angles into degrees. [Use π = 3.142] TP 2 Mempamerkan kefahaman tentang sukatan membulat. 1 0.81 rad 0.81 rad = 0.81 rad × 180° π = 46.40° 2 3.172 rad 3.172 rad = 3.172 rad × 180° π = 181.72° 3 5.79 rad 5.79 rad = 5.79 rad × 180° π = 331.70° 4 3.751 rad 3.751 rad = 3.751 rad × 180° π = 214.89° 5 2 7 π rad 2 7 π rad = 2 7 π rad × 180° π = 51.43° 6 3 17 π rad 3 17 π rad = 3 17 π rad × 180° π = 31.76° 7 2 2 17 π rad 2 2 17 π rad = 2 2 17 π rad × 180° π = 381.18° Latihan 2 Tukar setiap sudut yang berikut kepada radian. [Guna π = 3.142] TP 2 Convert each of the following angles into radians. [Use π = 3.142] TP 2 Mempamerkan kefahaman tentang sukatan membulat. 1 18° 18° = 18° × π 180° = 0.3142 rad 2 31.5° 31.5° = 31.5° × π 180° = 0.5499 rad 3 64.45° 64.45° = 64.45° × π 180° = 1.125 rad 4 128°26' 128°26′ = 128°26′ × π 180° = 2.242 rad 5 196° 196° = 196° × π 180° = 3.421 rad 88° Penyelesaian 88° = 88° × π 180° = 1.536 rad 88° Penyelesaian 88° = 88° × π 180° = 1.536 rad Contoh 2 Contoh 1 (a) 4.56 rad (b) 5 7 π rad Penyelesaian (a) 4.56 rad = 4.56 × 180° π = 261.23° (b) 5 7 π rad = 5 7 π × 180° π = 128.57° Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 2 30/10/2023 8:45:20 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

3 Tip 6 193.27° 193.27° = 193.27° × π 180° = 3.373 rad 7 273°31' 273°31′ = 273°31′ × π 180° = 4.774 rad 8 248° 248° = 248° × π 180° = 4.328 rad Tip Bestari Contoh 3 1.2 Panjang Lengkok Suatu Bulatan/ Arc Length of a Circle Latihan 3 Dengan menggunakan rumus s = jθ, cari panjang lengkok, s bagi setiap sektor yang berikut. [Guna π = 3.142] TP 2 By using the formula s = jθ, fi nd the arc length, s for each of the following sectors. [Use π = 3.142] TP 2 Mempamerkan kefahaman tentang sukatan membulat. 1 0.83 rad 4 cm s O s = jθ s = 4 × 0.83 = 3.32 cm 2 1.36 rad 6 cm O s s = jθ s = 6 × 1.36 = 8.16 cm 3 4.968 rad 8 cm s O s = jθ s = 8 × 4.968 = 39.74 cm j O s Q P fi s = jfi dengan keadaan, where, s = panjang lengkok arc length j = jejari/radius θ = sudut, dalam radian angle, in radian (a) 9 cm 78° O s Penyelesaian 78° = 78° × π 180° = 1.362 rad s = jθ = 9 × 1.362 = 12.26 cm (b) O s 7 cm 121° Penyelesaian 360° – 121° = 239° 239° = 239° × π 180° rad = 4.172 rad s = jθ = 7 × 4.172 = 29.20 cm Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 3 30/10/2023 8:45:21 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

4 Penyelesaian jθ = s j(0.986) = 15 j = 15 0.986 = 15.21 cm 0.986 rad O P Q 15 cm j cm Penyelesaian jθ = s j(0.986) = 15 j = 15 0.986 = 15.21 cm 0.986 rad O P Q 15 cm j cm 4 12 m s O 48° 48° = 48° × π 180° rad = 0.838 rad s = jθ = 12 × 0.838 = 10.06 m 5 16 cm s O 152° 152° = 152° × π 180° rad = 2.653 rad s = jθ = 16 × 2.653 = 42.45 cm 6 20 m s O 147° 360° – 147° = 213° 213° = 213° × π 180° rad = 3.718 rad s = jθ = 20 × 3.718 = 74.36 m Latihan 4 Dengan menggunakan rumus s = jθ, cari jejari, j. [Guna π = 3.142] TP 2 By using the formula s = jθ, fi nd the radius, j. [Use π = 3.142] TP 2 Mempamerkan kefahaman tentang sukatan membulat. 1 0.8 rad O P Q 12 cm j cm jθ = s j(0.8) = 12 j = 12 0.8 = 15 cm 2 0.845 rad O P Q 7 cm j cm jθ = s j(0.845) = 7 j = 7 0.845 = 8.284 cm 3 122.6° O P Q 16 cm j cm 122.6° = 122.6° × π 180° = 2.14 rad jθ = s j(2.14) = 16 j = 16 2.14 = 7.477 cm 4 3.96 rad O P Q 60 m j m jθ = s j(3.96) = 60 j = 60 3.96 = 15.15 m 5 351.75° P Q 75 cm j cm 351.75° = 351.75° × π 180° = 6.14 rad jθ = s j(6.14) = 75 j = 75 6.14 = 12.21 cm Contoh 4 Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 4 30/10/2023 8:45:22 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

5 Cari nilai θ, dalam radian. Find the value of θ, in radians. Penyelesaian jθ = s (6)θ = 8.5 θ = 8.5 6 = 1.417 rad O P Q 6 cm 8.5 cm θ Cari nilai θ, dalam radian. Find the value of θ, in radians. Penyelesaian jθ = s (6)θ = 8.5 θ = 8.5 6 = 1.417 rad O P Q 6 cm 8.5 cm θ Contoh 5 Contoh 6 Latihan 5 Selesaikan setiap yang berikut. [Guna π = 3.142] TP 2 Solve each of the following. [Use π = 3.142] TP 2 Mempamerkan kefahaman tentang sukatan membulat. 1 4 cm O P Q 5 cm θ Cari nilai θ, dalam radian. Find the value of θ, in radians. jθ = s (5)θ = 4 θ = 4 5 = 0.8 rad 2 O P Q 7.32 cm 6 cm θ Cari nilai θ, dalam radian. Find the value of θ, in radians. jθ = s (6)θ = 7.32 θ = 7.32 6 = 1.22 rad 3 12.8 cm O x x° P Q 7 cm Cari nilai x, dalam darjah. Find the value of x, in degrees. jθ = s (7)x = 12.8 x = 12.8 7 = 1.829 rad x = 1.829 × 180° π = 104.78° 4 O P Q 19 cm 8 cm O x° x P Q 19 cm 8 cm Cari nilai x, dalam darjah. Find the value of x, in degrees. jθ = s (8)x = 19 x = 19 8 = 2.375 rad x = 2.375 × 180° π = 136.06° 5 68 cm P Q O x x° 12 cm Cari nilai x, dalam darjah. Find the value of x, in degrees. jθ = s (12)x = 68 x = 68 12 = 5.667 rad x = 5.667 × 180° π = 324.65° Latihan 6 Hitung perimeter tembereng berlorek berikut. [Guna π = 3.142] TP 3 Calculate the perimeter of the following shaded segments. [Use π = 3.142] TP 3 Mengaplikasikan kefahaman tentang sukatan membulat untuk melaksanakan tugasan mudah. 1 15 cm R O Q P 0.8 rad 0.8 rad = 0.8 × 180° π = 45.83° Perimeter tembereng berlorek Perimeter of the shaded segment = 2(15)2 – 2(15)2 kos/cos 45.83° + (15)(0.8) = 11.681 + 12 = 23.68 cm R Q O P 9 cm 1.5 rad Penyelesaian 1.5 rad = 1.5 × 180° π = 85.93° Perimeter tembereng berlorek Perimeter of the shaded segment = 2j 2 – 2j 2 kos θ/cos θ + jθ = 2(9)2 – 2(9)2 kos 85.93°/cos 85.93° + (9)(1.5) = 12.268 + 13.5 = 25.77 cm Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 5 30/10/2023 8:45:23 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

6 2 9 cm 1.66 rad R O Q P 7 cm R Q O P 1.66 rad 1.66 rad = 1.66 × 180° π = 95.10° Perimeter tembereng berlorek Perimeter of the shaded segment = 2(7)2 – 2(7)2 kos/cos 95.10° + (7)(1.66) = 10.33 + 11.62 = 21.95 cm 3 R Q O P 1.74 rad 12 cm 1.74 rad = 1.74 × 180° π = 99.68° Perimeter tembereng berlorek Perimeter of the shaded segment = 2(12)2 – 2(12)2 kos/cos 99.68° + (12)(1.74) = 18.342 + 20.88 = 39.22 cm 4 R Q O P 2.56 rad 16 cm 2.56 rad = 2.56 × 180° π = 146.66° Perimeter tembereng berlorek Perimeter of the shaded segment = 2(16)2 – 2(16)2 kos/cos 146.66° + (16)(2.56) = 30.655 + 40.96 = 71.62 cm 5 R Q O P 15 cm 1.86 rad 1.86 rad = 1.86 × 180° π = 106.56° Perimeter tembereng berlorek Perimeter of the shaded segment = 2(15)2 – 2(15)2 kos/cos 106.56° + (15)(2π – 1.86) = 24.047 + 66.36 = 90.41 cm Latihan 7 Selesaikan setiap yang berikut. [Guna π = 3.142] TP 4 Solve each of the following. [Use π = 3.142] TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sukatan membulat dalam konteks penyelesaian masalah rutin yang mudah. 1 Rajah di bawah menunjukkan sebuah taman berbentuk sektor. The diagram below shows a garden in the shape of a sector. N O M 16 m 2.265 rad Taman itu akan dipagar. Hitung panjang, dalam m, pagar yang diperlukan. The garden has to be fenced. Calculate the length, in m, of the fence needed. Panjang pagar Length of the fence = Panjang lengkok MN + OM + ON Arc length MN + OM + ON = (16)(2.265) + 16 + 16 = 36.24 + 32 = 68.24 m Rajah di bawah menunjukkan sebuah kawasan bagi suatu pameran. The diagram below shows the area for an exhibition. G H O F 16 m 1.35 rad Hitung perimeter, dalam m, kawasan pameran itu. Calculate the perimeter, in m, of the exhibition area. Penyelesaian FOH = 1.35 rad = 1.35 rad × 180° π = 77.34° Perimeter = FH + Panjang lengkok FGH/ Arc length FGH = 2(16)2 – 2(16)2 kos/cos 77.34° + (16)(2π – 1.35) = 19.995 + 78.944 = 98.94 m Rajah di bawah menunjukkan sebuah kawasan bagi suatu pameran. The diagram below shows the area for an exhibition. G H O F 16 m 1.35 rad Hitung perimeter, dalam m, kawasan pameran itu. Calculate the perimeter, in m, of the exhibition area. Penyelesaian FOH = 1.35 rad = 1.35 rad × 180° π = 77.34° Perimeter = FH + Panjang lengkok FGH/ Arc length FGH = 2(16)2 – 2(16)2 kos/cos 77.34° + (16)(2π – 1.35) = 19.995 + 78.944 = 98.94 m Contoh 7 Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 6 30/10/2023 8:45:24 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

7 2 Rajah di bawah menunjukkan sebuah pentas berbentuk sektor berjejari 4 m. The diagram below shows a stage in the shape of a sector with a radius of 4 m. O 4 m 0.874 rad Sekeliling panjang lengkok pentas itu akan dihiasi dengan reben. Hitung panjang, dalam m, reben yang diperlukan. The curving part of the stage will be decorated with ribbons. Calculate the length, in m, of the ribbons needed. Panjang reben Length of ribbons = (4)(2π – 0.874) = (4)(5.41) = 21.64 m 3 Rajah di bawah menunjukkan pelan bagi sebuah tempat peranginan yang akan dibina. AOB ialah sektor berpusat O. The diagram below shows a plan of a resort that to be built. AOB is a sector with centre O. B O A 1.65 km Diberi bahawa perimeter tempat peranginan itu ialah 7.425 km. Cari AOB, dalam radian. It is given that the perimeter of the resort is 7.425 km. Find AOB, in radians. OA + OB + Panjang lengkok AB = 7.425 OA + OB + Arc length AB = 7.425 1.65 + 1.65 + (1.65)(/AOB) = 7.425 (1.65)(/AOB) = 4.125 /AOB = 2.5 radian/radians 4 Kawasan berlorek dalam rajah di bawah menunjukkan kawasan halaman rumput. POR ialah sektor berpusat O. The shaded region in the diagram below shows a lawn. POR is a sector with centre O. Hitung perimeter, dalam m, bagi kawasan halaman rumput. Calculate the perimeter, in m, of the lawn. kos/cos θ = 4 6 θ = 48.19° = 48.19° × π 180° = 0.8412 rad PQ = 6 × sin 48.19° = 4.472 m Perimeter = PQ + QR + Panjang lengkok PR/Arc length PR = 4.472 + 2 + (6)(0.8412) = 11.52 m 5 Dalam rajah di bawah, kawasan berlorek ialah sebuah pentas berbentuk tembereng berpusat O. In the diagram below, the shaded region is a stage in the shape of a segment with centre O. R O P Q 9 m 2.15 rad R O P Q 9 m 2.15 rad Sekeliling pentas itu akan diikat dengan reben. Hitung panjang, dalam m, reben yang diperlukan. The stage is to be surrounded by ribbons. Calculate the length, in m, of the ribbon needed. 2.15 rad = 2.15 × 180° π = 123.17° Panjang reben/ Length of ribbon = 2(9)2 – 2(9)2 kos/cos 123.17° + (9)(2.15) = 15.831 + 19.35 = 35.18 m Q O P R 6 m 4 m θ Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 7 30/10/2023 8:45:25 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

8 1.3 Luas Sektor Suatu Bulatan / Area of Sector of a Circle Latihan 8 Cari luas bagi sektor berlorek yang berikut. [Guna π = 3.142] TP 2 Find the area of the following shaded sectors. [Use π = 3.142] TP 2 Mempamerkan kefahaman tentang sukatan membulat. 1 Q O P 7 cm 0.85 rad Luas sektor OPQ Area of sector OPQ = 1 2 j 2 θ = 1 2 × 72 × 0.85 = 20.83 cm2 2 P O Q 6 cm 78° ∠POQ = 78° = 78° × π 180° = 1.362 rad Luas sektor OPQ Area of sector OPQ = 1 2 j 2 θ = 1 2 × 62 × 1.362 = 24.52 cm2 3 O P Q 9 cm 8 3 5 π Luas sektor OPQ Area of sector OPQ = 1 2 j 2 θ = 1 2 × 92 × 5 3 π = 212.09 cm2 4 P O Q 14 cm 160° ∠POQ = 360° – 160° = 200° 200° = 200° × π 180° = 3.491 rad Luas sektor OPQ Area of sector OPQ = 1 2 j 2 θ = 1 2 × 142 × 3.491 = 342.12 cm2 Contoh 8 Q O P 9 cm 83° Penyelesaian ∠POQ = 83° = 83° × π 180° = 1.449 rad Luas sektor OPQ Area of sector OPQ = 1 2 j 2 θ = 1 2 × 92 × 1.449 = 58.68 cm2 Kesilapan Umum (a) Luas sektor OPQ Area of sector OPQ = 1 2 × 92 × 83° = 3 361.5 cm2 θ mesti ditukarkan kepada radian. θ must be converted to radians. (b) Luas sektor OPQ Area of sector OPQ = 1 2 × 92 × 1.4 = 56.7 cm2 Nilai θ mestilah sekurangkurangnya dalam 2 tempat perpuluhan. The value of θ must be in at least 2 decimal places. Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 8 30/10/2023 8:45:25 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

9 Latihan 9 Diberi luas sektor berlorek POQ. Cari jejari bagi setiap yang berikut. [Guna π = 3.142] TP 2 Given the area of the shaded sector POQ. Find the radius of each of the following. [Use π = 3.142] TP 2 Mempamerkan kefahaman tentang sukatan membulat. 1 Luas/Area = 7.84 cm2 P O Q jr cm cm 0.684 rad 1 2 × j 2 × (0.684) = 7.84 j 2 = 7.84 × 2 0.684 = 22.924 j = 4.788 cm 2 Luas/Area = 116 cm2 P O Q r cm j cm 1.486 rad 1 2 × j 2 × (1.486) = 116 j 2 = 116 × 2 1.486 = 156.124 j = 12.49 cm 3 Luas/Area = 378 cm2 P O Q j cm 1.034 rad 1 2 × j 2 × (2π − 1.034) = 378 1 2 × j 2 × 5.25 = 378 j 2 = 378 × 2 5.25 = 144 j = 12 cm 4 Luas/Area = 471.28 cm2 1 2 × j 2 × (2π − 2.095) = 471.28 1 2 × j 2 × 4.189 = 471.28 j 2 = 471.28 × 2 4.189 = 225 j = 15 cm 5 Luas/Area = 535.43 cm2 47° × π 180° = 0.8204 rad 1 2 × j 2 × (2π − 0.8204) = 535.43 1 2 × j 2 × 5.4636 = 535.43 j 2 = 535.43 × 2 5.4636 = 196 j = 14 cm Luas/Area = 36 cm2 P O Q 1.2 rad j cm Penyelesaian 1 2 j 2 θ = Luas sektor POQ Area of sector POQ 1 2 × j 2 × (1.2) = 36 j 2 = 36 × 2 1.2 = 60 j = 60 = 7.746 cm P O Q j cm 47° P O Q j cm 2.095 rad Luas/Area = 36 cm2 P O Q 1.2 rad j cm Penyelesaian 1 2 j 2 θ = Luas sektor POQ Area of sector POQ 1 2 × j 2 × (1.2) = 36 j 2 = 36 × 2 1.2 = 60 j = 60 = 7.746 cm Contoh 9 Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 9 30/10/2023 8:45:26 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

10 Latihan 10 Diberi luas sektor berlorek POQ. Cari nilai θ, dalam radian, bagi setiap yang berikut. [Guna π = 3.142] TP 2 Given the area of the shaded sector POQ. Find the value of θ, in radians, for each of the following. [Use π = 3.142] TP 2 Mempamerkan kefahaman tentang sukatan membulat. 1 Luas/Area = 6.982 cm2 P O Q θ 4 cm 8 1 2 × 42 × θ = 6.982 θ = 6.982 × 2 42 θ = 0.8728 rad 2 Luas/Area = 102.8 cm2 P O Q 12 cm θ 1 2 × 122 × θ = 102.8 θ = 102.8 × 2 122 θ = 1.428 rad 3 Luas/Area = 128.30 cm2 P O Q 7 cm θ 1 2 × 72 × θ = 128.30 θ = 128.30 × 2 72 θ = 5.237 rad 4 Luas/Area = 240 cm2 P O Q 14 cm θ 1 2 × 142 × (2π − θ) = 240 2π − θ = 2.449 θ = 2π − 2.449 = 3.835 rad 5 Luas/Area = 693.76 cm2 P O Q 16 cm θ 1 2 × 162 × (2π − θ) = 693.76 2π − θ = 5.42 θ = 2π − 5.42 = 0.864 rad Luas/Area = 39.872 cm2 P O Q 8 cm θ Penyelesaian 1 2 j 2 θ = Luas sektor POQ Area of sector POQ 1 2 × 82 × θ = 39.872 θ = 39.872 × 2 82 = 1.246 rad Luas/Area = 39.872 cm2 P O Q 8 cm θ Penyelesaian 1 2 j 2 θ = Luas sektor POQ Area of sector POQ 1 2 × 82 × θ = 39.872 θ = 39.872 × 2 82 = 1.246 rad Contoh 10 Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 10 30/10/2023 8:45:26 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

11 Latihan 11 Cari luas tembereng berlorek bagi setiap yang berikut. [Guna π = 3.142] TP 3 Find the area of the shaded segment of each of the following. [Use π = 3.142] TP 3 Mengaplikasikan kefahaman tentang sukatan membulat untuk melaksanakan tugasan mudah 1 P O Q 12 cm 3.1 rad 3.1 rad = 3.1 × 180° π = 177.59° Luas tembereng berlorek/ Area of the shaded segment = 1 2 × j 2 × θ × − 1 2 × j 2 × sin θ = 1 2 × 122 × 3.1 − 1 2 × 122 × sin 177.59° = 223.2 – 3.028 = 220.17 cm2 2 P O Q 16 cm 68° 68° = 68° × π 180° = 1.187 rad Luas tembereng berlorek/ Area of the shaded segment = 1 2 × j 2 × θ − 1 2 × j 2 × sin θ = 1 2 × (16)2 × 1.187 − 1 2 × (16)2 × sin 68° = 151.936 − 118.680 = 33.26 cm2 3 P O Q 15 cm 136° 136° = 136° × π 180° = 2.374 rad Luas tembereng berlorek/Area of the shaded segment = 1 2 × j 2 × θ − 1 2 × j 2 × sin θ = 1 2 × (15)2 × 2.374 − 1 2 × (15)2 × sin 136° = 267.075 − 78.149 = 188.93 cm2 P O Q 12 cm 1.45 rad Penyelesaian 1.45 rad = 1.45 × 180° π = 83.07° Luas tembereng berlorek/ Area of the shaded segment = 1 2 × j 2 × θ − 1 2 × j 2 × sin θ = 1 2 × (12)2 × 1.45 − 1 2 × (12)2 × sin 83.07° = 104.4 − 71.47 = 32.93 cm2 Contoh 11 Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 11 30/10/2023 8:45:27 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

12 Latihan 12 Selesaikan masalah yang berikut. TP 4 Solve the following problems. TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sukatan membulat dalam konteks penyelesaian masalah rutin yang mudah. 1 Rajah di bawah menunjukkan sebuah semibulatan berdiameter 16 cm. The diagram below shows a semicircle with a diameter of 16 cm. C A B 0.68 rad 16 cm Cari luas, dalam cm2 , kawasan berlorek. Find the area, in cm2 , of the shaded region. [Guna/Use π = 3.142] Katakan O ialah pusat semibulatan. Let O be the centre of the semicircle. ∠COB = 2 × ∠CAO 2 × 0.68 = 1.36 rad π − 1.36 rad = 1.782 rad 1.782 × 180o π = 102.09o C A B 0.68 rad 8 cm 1.36 rad 1.782 rad O Luas tembereng berlorek Area of the shaded segment = Luas ∆AOC + Luas sektor COB Area of ∆AOC + Area of sector COB = 1 2 × 82 × sin 102.09o + 1 2 × 82 × 1.36 = 31.29 + 43.52 = 74.81 cm2 2 Rajah di bawah menunjukkan pelan bagi sebuah taman. HQG ialah sebuah semibulatan berpusat O dan berjejari 10 m. FPG ialah sebuah sektor berpusat P dan berjejari 18 m. Sektor QOG ialah kawasan halaman rumput. Kawasan berlorek ialah kawasan tanaman bunga. Seluruh kawasan tanaman bunga akan dipagarkan. Diberi bahawa PQ = 10 m dan QOG = 1.795 rad. The diagram below shows the plan of a garden. HQG is a semicircle with centre O and a radius of 10 m. FPG is a sector with centre P and a radius of 18 m. Sector QOG is a lawn. The shaded region is a fl ower bed. The whole fl ower bed has to be fenced. It is given that PQ = 10 m and QOG = 1.795 rad. Hitung/Calculate KBAT Menilai (a) luas, dalam m2 , kawasan halaman rumput, the area, in m2 , of the lawn, (b) panjang, dalam m, pagar yang diperlukan, the length, in m, of the fence needed, (c) luas, dalam m2 , kawasan tanaman bunga. the area, in m2 , of the fl ower bed. [Guna/Use π = 3.142] ∠FPO = ∠QOP = π – 1.795 rad = 1.347 rad = 1.347 × 180o π = 77.17o (a) 1 2 × 102 × 1.795 = 89.75 m2 (b) Panjang pagar/ Length of the fence = QF + Panjang lengkok FG/Arc length FG + Panjang lengkok QG/Arc length QG = 8 + 18(1.347) + 10(1.795) = 50.20 m (c) Luas tanaman bunga/ Area of fl ower bed = Luas sektor FPG – Luas ∆QPO – Luas sektor QOG Area of sector FPG − Area of ∆QPO – Area of sector QOG = ( 1 2 × 182 × 1.347) − ( 1 2 × 10 × 8 sin 77.17o ) − ( 1 2 × 102 × 1.795) = 218.21 − 39.001 − 89.75 = 89.46 m2 F Q H P O G Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 12 30/10/2023 8:45:27 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

13 1.4 Aplikasi Sukatan Membulat/ Application of Circular Measures Latihan 13 Selesaikan masalah yang berikut. TP 4 Solve the following problems. TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang sukatan membulat dalam konteks penyelesaian masalah rutin yang mudah. 1 Rajah di sebelah menunjukkan pelan bagi sebuah taman bunga yang berbentuk semibulatan berpusat O dan diameter 24 m. Kawasan berlorek akan ditanam pokok bunga ros. Panjang lengkok KL adalah sama dengan jumlah panjang OL dan panjang lengkok LM. The diagram on the right shows the plan of a fl ower garden in the shape of a semicircle with centre O and a diameter of 24 m. The shaded region is to be planted with rose plants. The arc length KL is equal to the total of the length OL and the arc length LM. Jika setiap keluasan 2 m2 boleh ditanam dengan 5 pokok bunga ros, cari bilangan pokok bunga ros yang boleh ditanam dalam kawasan berlorek. If each area of 2 m2 can be planted with 5 rose plants, fi nd the number of rose plants that can be planted in the shaded region. Jejari/ Radius = 24 m ÷ 2 = 12 m Panjang lengkok KL = OL + Panjang lengkok LM Arc length KL = OL + Arc length LM 12 × KOL = 12 + 12(π – KOL) KOL = 1 + π – KOL 2KOL = 1 + π KOL = 1 2 (1 + π) = 2.071 rad Luas sektor KOL/ Area sector KOL = 1 2 × 2.071 × 122 = 149.11 m2 2 Rajah di bawah menunjukkan pelan bagi sebuah taman. EOF dan GOH ialah dua buah sektor dengan pusat sepunya O. Diberi GO = 20 m, GE = EO dan panjang lengkok GH = 8π m. The diagram below shows the plan of a park. EOF and GOH are two sectors with common centre O. Given GO = 20 m, GE = EO and the arc length GH = 8π m. Kawasan berlorek akan ditutupi dengan jubin. Jika kos untuk menutup kawasan seluas 1 m2 ialah RM150, hitung jumlah kos untuk menutup kawasan berlorek itu. The shaded region will be covered with tiles. If the cost to cover an area of 1 m2 is RM150, calculate the total cost to cover the shaded region. Panjang lengkok GH/ Arc length GH = GO × GOH 8π = 20 × GOH GOH = 2π 5 rad Luas kawasan berlorek/ Area of the shaded region = Luas sektor GOH – Luas sektor EOF/ Area of sector GOH – Area of sector EOF = 1 2 × 202 × 2π 5 – 1 2 × 102 × 2π 5 = 251.36 – 62.84 = 188.52 m2 Jumlah kos/Total cost = 188.52 × RM150 = RM28 278 Luas kawasan yang akan ditanam pokok bunga ros ialah 149.11 m2 . Diberi 2 m2 boleh ditanam 5 pokok bunga ros. Maka, 1 m2 boleh ditanam 5 2 pokok bunga ros. Area to be planted with rose plants is 149.11 m2 . Given that 2 m2 can be planted with 5 rose plants. Thus, 1 m2 can be planted with 5 2 rose plants. Bilangan pokok bunga ros yang boleh ditanam Number of rose plants to be planted = 149.11 × 5 2 = 372.78 372 pokok bunga ros/ rose plants G H O E F KBAT Mengaplikasi Untuk tujuan pembelajaran Imbas kod QR atau layari htt ps://youtu.be/O-- AWE3BoEY untuk menonton video tutorial penyelesaian masalah yang melibatkan luas sektor. Video Tutorial L Dinding bangunan Wall of building M O K Video Tutorial Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 13 30/10/2023 8:45:27 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

14 Kertas 1 Bahagian A 1 Rajah 1 menunjukkan sektor AOB berpusat O dan berjejari 12 cm. Diagram 1 shows a sector AOB with centre O and a radius of 12 cm. A B O 12 cm Rajah 1 / Diagram 1 Diberi bahawa panjang lengkok AB ialah 25.8 cm. Cari luas, dalam cm2 , sektor AOB. It is given that the arc length AB is 25.8 cm. Find the area, in cm2 , of the sector AOB. [3 markah/marks] 2 Rajah 2 menunjukkan sektor FOG yang mencangkum sudut 5 12 π radian pada pusat bulatan, O. Diberi bahawa jejari bulatan itu ialah 6 cm. Diagram 2 shows a sector FOG which subtends an angle of 5 12 π radians at the centre of a circle, O. It is given that the radius of the circle is 6 cm. F O G 5π 12 rad Rajah 2 / Diagram 2 Cari dalam sebutan π, Find in terms of π, (a) panjang lengkok major FG, dalam cm, the major arc length FG, in cm, (b) luas sektor major FOG, dalam cm2 . the area of the major sector FOG, in cm2 . [4 markah/marks] 3 Rajah 3 menunjukkan sebuah semibulatan berpusat O dan berjejari 8 cm. Diagram 3 shows a semicircle with centre O and a radius of 8 cm. Q θ P R O Rajah 3 / Diagram 3 Diberi panjang lengkok PQ adalah sama dengan perimeter sektor QOR, hitung Given that the arc length PQ is equal to the perimeter of the sector QOR, calculate (a) nilai θ, dalam radian, the value of θ, in radians, (b) luas sektor QOR, dalam cm². the area of sector QOR, in cm². [Guna/ Use π = 3.142] [4 markah/marks] 4 Rajah 4 menunjukkan sebuah bulatan berpusat O dan berjejari 8 cm. Diagram 4 shows a circle with centre O and a radius of 8 cm. O R P Q 8 cm θ 2.5 cm Rajah 4 / Diagram 4 Diberi PQ = 2.5 cm, cari Given that PQ = 2.5 cm, find (a) nilai θ, dalam radian, the value of θ, in radians, (b) luas kawasan berlorek, dalam cm². the area of the shaded region, in cm². [Guna/Use π = 3.142] [5 markah/marks] 5 Dalam Rajah 5, A, B, C dan D ialah titik-titik pada lilitan sebuah bulatan berpusat O dan berjejari 17 m. Kawasan berlorek ialah sebuah jambatan. In Diagram 5, A, B, C and D are points on the circumference of a circle with centre O and a radius of 17 m. The shaded region is a bridge. O B P Q D C A30 m 30 m 17 m θ Rajah 5 / Diagram 5 Hitung/Calculate (a) nilai θ, dalam radian, the value of θ, in radians, (b) luas tembereng BQC, dalam m², the area of the segment BQC, in m², (c) luas, dalam m², jambatan itu. the area, in m², of the bridge. [Guna/Use π = 3.142] [6 markah/marks] Praktis Berformat SPM Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 14 30/10/2023 8:45:27 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

15 Kertas 2 Bahagian A 1 Rajah 1 menunjukkan sebuah trapezium EJKL. G ialah titik tengah LK. EFG ialah lengkok bagi sebuah sektor berpusat O. Diagram 1 shows a trapezium EJKL. G is the midpoint of LK. EFG is an arc of a sector with centre O. F O L G K E J 30 cm 24 cm 5 cm Rajah 1 / Diagram 1 Diberi LK = 24 cm dan OL = 5 cm, hitung Given that LK = 24 cm and OL = 5 cm, calculate (a) EOG, dalam radian, EOG, in radians, [1 markah/mark] (b) perimeter, dalam cm, bagi kawasan berlorek, the perimeter, in cm, of the shaded region, [3 markah/marks] (c) luas, dalam cm², bagi kawasan berlorek. the area, in cm2 , of the shaded region. [3 markah/marks] 2 Rajah 2 menunjukkan sebuah bulatan berpusat O. Diagram 2 shows a circle with centre O. P S R Q θ O Rajah 2 / Diagram 2 Diberi bahawa panjang lengkok PQR ialah 31.5 cm dan luas kawasan berlorek ialah 118 1 8 cm2 . It is given that the arc length PQR is 31.5 cm and the area of the shaded region is 118 1 8 cm2 . [Guna/Use π = 3.142] Hitung/Calculate (a) jejari, dalam cm, bulatan itu, the radius, in cm, of the circle, [3 markah/marks] (b) nilai θ, dalam radian, the value of θ, in radians, [2 markah/marks] (c) luas, dalam cm2 , bagi tembereng PRS. the area, in cm2 , of the segment PRS. [3 markah/marks] 3 Rajah 3 menunjukkan sebuah semibulatan berpusat O dan berdiameter 26 cm. BA ialah tangen kepada semibulatan di titik A. Diagram 3 shows a semicircle with centre O and a diameter of 26 cm. BA is a tangent to the semicircle at point A. A C D B O Rajah 3 / Diagram 3 Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 15 30/10/2023 8:45:28 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

16 Diberi bahawa panjang lengkok CD ialah 27.3 cm. It is given that the arc length CD is 27.3 cm. [Guna/Use π = 3.142] Hitung/Calculate (a) COD, dalam radian, COD, in radians, [3 markah/marks] (b) luas, dalam cm2 , bagi kawasan berlorek. the area, in cm2 , of the shaded region. [4 markah/marks] Bahagian B 4 Rajah 4 menunjukkan sebuah bulatan berpusat O dan berjejari 18 cm terterap dalam sektor FEG berpusat E. Garis lurus EF dan EG ialah tangen kepada bulatan masing-masing di titik H dan titik K. Diagram 4 shows a circle with centre O and a radius of 18 cm inscribed in a sector FEG with centre E. The straight lines EF and EG are tangents to the circle at point H and point K respectively. 60˚ F P G K E O H 18 cm Rajah 4 / Diagram 4 Hitung/Calculate (a) panjang lengkok FG, dalam cm, the arc length FG, in cm, [5 markah/marks] (b) luas, dalam cm2 , bagi kawasan berlorek. the area, in cm2 , of the shaded region. [5 markah/marks] Zon KBAT 1 Rajah 1 menunjukkan dua buah bulatan masing-masing berpusat K dan L yang bercantum di titik P. Bulatan yang lebih besar mempunyai jejari 14 cm dan bulatan yang lebih kecil mempunyai jejari 10 cm. Garis lurus EFGH ialah tangen sepunya kepada kedua-dua bulatan itu. Diagram 1 shows two circles with centres K and L respectively and touch at point P. The larger circle has a radius of 14 cm and the smaller circle has a radius of 10 cm. The straight line EFGH is a common tangent to both circles. E F G H K L P θ 10 cm 14 cm Rajah 1 / Diagram 1 Diberi bahawa KLG = θ radian. It is given that KLG = θ radians. (a) Tunjukkan bahawa θ = 1.404 (betul kepada tiga tempat perpuluhan). KBAT Mengaplikasi Show that θ = 1.404 (correct to three decimal places). (b) Hitung panjang lengkok minor FP, dalam cm. Calculate the minor arc length FP, in cm. (c) Hitung luas, dalam cm2 , bagi kawasan berlorek. KBAT Menilai Calculate the area, in cm2 , of the shaded region. Strategi A+ Maths Tam Tg5-01_vim_3p(1-16).indd 16 30/10/2023 8:45:28 AM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

17 Jika y = g(u) dan u = h(x), maka dy dx = dy du × du dx If y = g(u) and u = h(x), then dy dx = dy du × du dx 4 Terbitan pertama bagi suatu fungsi yang melibatkan pendaraban dua ungkapan algebra boleh diperoleh dengan menggunakan petua hasil darab. The fi rst derivative of a function involving product of two algebraic expressions can be obtained by using product rule. Jika y = uv, dengan keadaan u = f(x) dan v = g(x), maka dy dx = u dv dx + v du dx If y = uv, where u = f(x) and v = g(x), then dy dx = u dv dx + v du dx 5 Terbitan pertama bagi suatu fungsi yang melibatkan pembahagian dua ungkapan algebra boleh diperoleh dengan menggunakan petua hasil bahagi. The fi rst derivative of a function involving quotient of two algebraic expressions can be obtained by using quotient rule. Jika y = u v , dengan keadaan u = f (x) dan v = g(x), maka dy dx = v du dx – u dv dx v2 If y = u v , where u = f (x) and v = g(x), then dy dx = v du dx – u dv dx v2 2.3 Pembezaan Peringkat Kedua The Second Derivative 1 Terbitan kedua bagi suatu fungsi y = ƒ(x) diperoleh dengan membezakan fungsi itu sebanyak dua kali berturut-turut dan ditulis sebagai d2 y dx2 atau ƒ''(x). The second derivative of a function y = f(x) is obtained by differentiating the function through two consecutive times and written as d2 y dx2 or f''(x). d2 y dx2 = d dx ( dy dx ) atau/ or ƒ'' (x) = d dx [ƒ'(x)] Contoh/ Example y = 3x3 + x2 – 7 dy dx = 9x2 + 2x Pembezaan peringkat pertama First derivative d2 y dx2 = 18x + 2 Pembezaan peringkat kedua Second derivative 2.1 Had dan Hubungannya dengan Pembezaan Limit and its Relation to Differentiation 1 had ƒ(x) = L dibaca sebagai “had bagi ƒ(x) apabila x x " a menghampiri a ialah L”. lim f(x) = L is read as “the limit of ƒ(x) as x approaches a is L”. x " a 2 Terbitan pertama bagi suatu fungsi y = ƒ(x) ditentukan menggunakan kaedah pembezaan prinsip pertama. The fi rst derivative of a function y = ƒ(x) is determined using differentiation using fi rst principle. dy dx = had/lim ƒ(x + δx) – ƒ(x) δx " 0 δx " 0 δx 2.2 Pembezaan Peringkat Pertama The First Derivative 1 Terbitan pertama bagi fungsi y = ƒ(x) ditulis sebagai dy dx atau ƒ'(x). The fi rst derivative of a function y = ƒ(x) is written as dy dx or ƒ'(x). (a) Jika y = a, maka dy dx = 0, dengan keadaan a ialah pemalar. If y = a, then dy dx = 0, where a is a constant. (b) Jika y = ax, maka dy dx = a, dengan keadaan a ialah pemalar. If y = ax, then dy dx = a, where a is a constant. (c) Jika y = axn , maka dy dx = anxn – 1, dengan keadaan a dan n ialah pemalar. If y = axn , then dy dx = anxn – 1, where a and n are constants. 2 Terbitan pertama bagi suatu fungsi algebra yang melibatkan penambahan dan penolakan boleh ditentukan dengan membezakan setiap sebutan secara berasingan. The fi rst derivative of an algebraic function involving addition and subtraction can be determined by differentiating each term separately. d dx [ƒ(x) ± g(x)] = d dx [ƒ(x)] ± d dx [g(x)] Contoh/ Example y = x6 + 4x d dx (x6 + 4x) = d dx (x6 ) + d dx (4x) = 6x5 + 4 3 Terbitan pertama bagi suatu fungsi gubahan boleh diperoleh dengan menggunakan petua rantai. The fi rst derivative of a composite function can be obtained by using chain rule. Revisi Pantas Bab 2 Pembezaan Differentiation Bidang Pembelajaran: Kalkulus Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 17 26/10/2023 12:12:03 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

18 2.4 Aplikasi Pembezaan Application of Differentiation 1 Garis lurus yang menyentuh satu titik pada lengkung dikenali sebagai tangen. A straight line that touches a point on a curve is known as a tangent. y Tangen Tangent 0 y = f(x) P(x1 , y1 ) x Normal y Tangen Tangent y = f(x) P(x1 , y1 ) m1 m2 x 0 l = 2 Kecerunan tangen berbeza bagi setiap titik yang berlainan pada suatu lengkung. The gradient of the tangent is different for each different point on a curve. 3 Kecerunan tangen, m1 boleh ditentukan dengan menggantikan nilai x1 ke dalam dy dx . The gradient of the tangent, m1 can be determined by substituting the value of x1 into dy dx . 4 Persamaan tangen pada titik P(x1 , y1 ) ialah y – y1 = m1 (x – x1 ). The equation of the tangent at the point P(x1 , y1 ) is y – y1 = m1 (x – x1 ). 5 Garis normal ialah garis yang berserenjang dengan tangen. Maka, kecerunan normal, m2 ialah – 1 dy dx . Normal line is a line that is perpendicular to the tangent. Hence, the gradient of the normal, m2 is – 1 dy dx . y Tangen Tangent 0 y = f(x) P(x1 , y1 ) x Normal y Tangen Tangent y = f(x) P(x1 , y1 ) m1 m2 x 0 l = 6 Persamaan normal pada titik P(x1 , y1 ) ialah y – y1 = m2 (x – x1 ). The equation of the normal at the point P(x1 , y1 ) is y – y1 = m2 (x – x1 ). 7 Titik pegun ialah titik pada suatu lengkung y = ƒ(x) dengan keadaan fungsi kecerunan pada titik itu ialah 0, dy dx = 0. Stationary point is a point on a curve y = f(x) such that the gradient function at the point is 0, dy dx = 0. 8 Terdapat tiga jenis titik pegun pada suatu lengkung: There are three types of stationary points on a curve: (a) titik maksimum/maximum point (b) titik minimum/minimum point (c) titik lengkok balas/point of infl ection 9 Diberi suatu lengkung y = ƒ(x) dengan titik pegun P di x = x1 . Given a curve y = f(x) with a stationary point P at x = x1 . (a) Jika tanda dy dx berubah daripada positif kepada negatif apabila x merentasi x1 dari kiri ke kanan graf, maka titik P ialah titik maksimum. If the sign of dy dx changes from positive to negative as x crosses x1 from left to right of the graph, then point P is a maximum point. dy dx = 0 dy dx < 0 dy dx > 0 x1 dy dx = 0 dy dx < 0 dy dx > 0 x1 dy dx = 0 dy dx > 0 x1 dy dx > 0 (b) Jika tanda dy dx berubah daripada negatif kepada positif apabila x merentasi x1 dari kiri ke kanan graf, maka titik P ialah titik minimum. If the sign of dy dx changes from negative to positive as x crosses x1 from left to right of the graph, then point P is a minimum point. dy dx = 0 dy dx < 0 dy dx > 0 x1 dy dx = 0 dy dx < 0 dy dx > 0 x1 dy dx = 0 dy dx > 0 x1 dy dx > 0 (c) Jika tanda dy dx tidak berubah daripada positif kepada negatif atau sebaliknya apabila x merentasi x1 , maka titik P ialah titik lengkok balas. If the sign of dy dx does not change as x crosses x1 from left to right of the graph, then point P is a point of infl ection. dy dx = 0 dy dx < 0 dy dx > 0 x1 dy dx = 0 dy dx < 0 dy dx > 0 x1 dy dx = 0 dy dx > 0 x1 dy dx > 0 10 Titik maksimum dan titik minimum juga dikenali sebagai titik pusingan. Maximum point and minimum point are also known as turning point. 11 Langkah-langkah untuk menentukan koordinat titik pusingan: Steps to determine the coordinates of the turning point: (a) Cari dy dx . / Find dy dx . (b) Selesaikan dy dx = 0 untuk mencari nilai x. Solve dy dx = 0 to fi nd the value of x. 12 Terdapat dua kaedah untuk menentukan sama ada suatu titik pusingan ialah titik maksimum atau minimum. There are two methods to determine whether a turning point is a maximum or minimum point. Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 18 26/10/2023 12:12:04 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

19 i-THINK Peta Pokok Lakaran tangen Sketching of tangents Pembezaan peringkat kedua Second order derivative • Bagi titik maksimum, nilai dy dx berubah daripada positif kepada sifar dan kemudian kepada negatif. For maximum point, the value of dy dx changes from positive to zero and then to negative. • Bagi titik minimum, nilai dy dx berubah daripada negatif kepada sifar dan kemudian kepada positif. For minimum point, the value of dy dx changes from negative to zero and then to positive. • Bagi titik maksimum, nilai d2 y dx2 ialah negatif. For maximum point, the value of d2 y dx2 is negative. • Bagi titik minimum, nilai d2 y dx2 ialah positif. For minimum point, the value of d2 y dx2 is positive. Kaedah menentukan titik pusingan Methods to determine a turning point 13 Jika dua pemboleh ubah, x dan y berubah terhadap masa, t dan dihubungkan oleh persamaan y = ƒ(x), maka kadar perubahan dx dt dan dy dt boleh dihubungkan oleh If two variables, x and y change with respect to time, t and are related by the equation y = f(x), then the rates of change dx dt and dy dt can be related by dy dt = dy dx × dx dt (a) dy dt > 0 menunjukkan kadar pertambahan y terhadap t. dy dt > 0 shows the rate of increase of y with respect to t. (b) dy dt < 0 menunjukkan kadar penyusutan y terhadap t. dy dt < 0 shows the rate of decrease of y with respect to t. 14 Perubahan hampir dalam y, δy akibat perubahan kecil dalam x diberi oleh The approximate change in y, δy caused by a small change in x is given by δy = dy dx × δx (a) Jika δy > 0, maka berlaku tokokan kecil dalam y akibat perubahan kecil dalam x, δx. If δy > 0, then there is a small increase in y due to a small change in x, δx. (b) Jika δy < 0, maka berlaku susutan kecil dalam y akibat perubahan kecil dalam x, δx. If δy < 0, then there is a small decrease in y due to a small change in x, δx. 15 Nilai hampir bagi y boleh ditentukan menggunakan rumus berikut. The approximate value of y can be determined using the following formula. ƒ(x + δx) ≈ y + dy dx δx 1 had / lim x → 2 x → 2 (2x2 – x) had / lim x → 2 x → 2 (2x2 – x) = 2(2)2 – (2) = 8 – 2 = 6 2 had / lim x → 2 x → 2 (x2 – 6x + 3) had / lim x → 2 x → 2 (x2 – 6x + 3) = (2)2 – 6(2) + 3 = 4 – 12 + 3 = –5 Praktis PBD 2.1 Had dan Hubungannya dengan Pembezaan/ Limit and its Relation to Differentiation Latihan 1 Nilaikan setiap yang berikut. TP 2 Evalute each of the following. TP 2 Mempamerkan kefahaman tentang pembezaan. had / lim x → 3 x → 3 (x2 + 5x – 2) Penyelesaian = had / lim x → 3 x → 3 (x2 + 5x – 2) = (3)2 + 5(3) – 2 = 9 + 15 – 2 = 22 Contoh 1 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 19 26/10/2023 12:12:04 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

20 Latihan 2 Nilaikan setiap yang berikut. TP 2 Evaluate each of the following. TP 2 Mempamerkan kefahaman tentang pembezaan. 1 had / lim x → –6 x → –6 36 – x2 x + 6 had / lim x → –6 x → –6 36 – x2 x + 6 = had / lim x → –6 x → –6 (6 + x)(6 – x) x + 6 = had / lim x → –6 x → –6 (6 – x) = 6 – (–6) = 12 2 had / lim x → –2 x → –2 x2 – x – 6 x + 2 had / lim x → –2 x → –2 x2 – x – 6 x + 2 = had / lim x → –2 x → –2 (x + 2)(x – 3) x + 2 = had / lim x → –2 x → –2 (x – 3) = (–2) – 3 = –5 3 had / lim x → –3 x → –3 15 + 2x – x2 x + 3 had / lim x → –3 x → –3 15 + 2x – x2 x + 3 = had / lim x → –3 x → –3 (3 + x)(5 – x) x + 3 = had / lim x → –3 x → –3 (5 – x) = 5 – (–3) = 8 4 had / lim x → 4 x → 4 x2 – 2x – 8 x2 – 16 had / lim x → 4 x → 4 x2 – 2x – 8 x2 – 16 = had / lim x → 4 x → 4 (x + 2)(x – 4) (x + 4)(x – 4) = had / lim x → 4 x → 4 x + 2 x + 4 = (4) + 2 (4) + 4 = 6 8 = 3 4 5 had / lim x → 3 x → 3 x2 + x – 12 x2 – 9 had / lim x → 3 x → 3 x2 + x – 12 x2 – 9 = had / lim x → 3 x → 3 (x – 3)(x + 4) (x + 3)(x – 3) = had / lim x → 3 x → 3 x + 4 x + 3 = (3) + 4 (3) + 3 = 7 6 3 had / lim x → 3 x → 3 4x 2x + 1 had / lim x → 3 x → 3 4x 2x + 1 = 4(3) 2(3) + 1 = 12 7 4 had / lim x → 4 x → 4 10 + x x + 3 had / lim x → 4 x → 4 10 + x x + 3 = 10 + 4 4 + 3 = 14 7 = 2 5 had / lim x → 4 x → 4 3x + 4 had / lim x → 4 x → 4 3x + 4 = 3(4) + 4 = 16 = 4 Contoh 2 had / lim x → 5 x → 5 3x2 – 75 x – 5 Penyelesaian had / lim x → 5 x → 5 3x2 – 75 x – 5 = had / lim x → 5 x → 5 3(x2 – 25) x – 5 = had / lim x → 5 x → 5 3(x + 5)(x – 5) x – 5 = had / lim x → 5 x → 5 3(x + 5) = 3(5 + 5) = 30 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 20 26/10/2023 12:12:06 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

21 Latihan 3 Nilaikan setiap yang berikut. TP 2 Evaluate each of the following. TP 2 Mempamerkan kefahaman tentang pembezaan. 1 had m → 3 /limm → 3 1 m − 3 m2 − 92 = had / lim m → 3 m → 3 m− 3 (m+ 3)(m− 3) = had / lim m → 3 m → 3 1 m + 3 = 1 (3) + 3 = 1 6 2 had m → 6 /limm → 6 1 m− 6 m2 − 362 = had / lim m → 6 m → 6 m− 6 (m+ 6)(m− 6) = had / lim m → 6 m → 6 1 m + 6 = 1 (6) + 6 = 1 12 3 had m → 5 /limm → 5 1 25 − m2 5 − m 2 = had / lim m → 5 m → 5 (5 + m)(5 − m) 5 − m = had / lim m → 5 m → 5 (5 + m) = 5 + (5) = 10 4 had m → 2 /limm → 2 1 2m2 − 8 m − 2 2 = had / lim m → 2 m → 2 2(m2 − 4) m − 2 = had / lim m → 2 m → 2 2(m+ 2)(m− 2) m– 2 = had / lim m → 2 m → 2 2(m+ 2) = 2[(2) + 2] = 8 5 had m → 1 /limm → 1 1 3m2 − 3 m − 1 2 = had / lim m → 1 m → 1 3(m2 − 1) m − 1 = had / lim m → 1 m → 1 3(m+ 1)(m− 1) m– 1 = had / lim m → 1 m → 1 3(m+ 1) = 3[(1) + 1] = 6 Latihan 4 Dengan menggunakan rumus dy dx = had δx → 0 f(x + δx) – f(x) δx , cari dy dx . TP 2 By using the formula dy dx = lim δx → 0 f(x + δx) – f(x) δx , fi nd dy dx . TP 2 Mempamerkan kefahaman tentang pembezaan. 1 y = 4x + 5 dy dx = had / lim δx → 0 δx → 0 f(x + δx) – f(x) δx = had / lim δx → 0 δx → 0 [4(x + δx) + 5] – (4x + 5) δx = had / lim δx → 0 δx → 0 4x + 4δx + 5 – 4x – 5 δx = had / lim δx → 0 δx → 0 4δx δx = had / lim δx → 0 δx → 0 4 = 4 had m → 2 /limm → 2 1 m2 – 4 m – 22 Penyelesaian had m → 2 /limm → 2 1 m2 – 4 m – 22 = had m → 2 /limm → 2 (m+ 2)(m– 2) m – 2 = had m → 2 /limm → 2 (m + 2) = (2) + 2 = 4 Contoh 3 Kesilapan Umum had m → 2 /limm → 2 m2 – 4 m – 2 = (2)2 – 4 (2) – 2 = 0 0 (Tidak tertakrif/Undefi ned) Contoh 4 y = 7x2 Penyelesaian dy dx = had δx → 0 / lim δx → 0 f(x + δx) – f(x) δx = had δx → 0 / lim δx → 0 7(x + δx)2 – 7x2 δx = had δx → 0 / lim δx → 0 7[x2 + 2xδx + (δx)2 ] – 7x2 δx = had δx → 0 / lim δx → 0 7x2 + 14xδx + 7(δx)2 – 7x2 δx = had δx → 0 / lim δx → 0 14xδx + 7(δx)2 δx = had δx → 0 / lim δx → 0 (14x + 7δx) = 14x + 7(0) = 14x Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 21 26/10/2023 12:12:08 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

22 2 y = 2x2 dy dx = had / lim δx → 0 δx → 0 f(x + δx) – f(x) δx = had / lim δx → 0 δx → 0 2(x + δx)2 – 2x2 δx = had / lim δx → 0 δx → 0 2[x2 + 2xδx + (δx)2 ] – 2x2 δx = had / lim δx → 0 δx → 0 2x2 + 4xδx + 2(δx)2 – 2x2 δx = had / lim δx → 0 δx → 0 4xδx + 2(δx)2 δx = had / lim δx → 0 δx → 0 (4x + 2δx) = 4x + 2(0) = 4x 3 y = x2 + x dy dx = had / lim δx → 0 δx → 0 f(x + δx) – f(x) δx = had / lim δx → 0 δx → 0 (x + δx) 2 + (x + δx) – (x 2 + x) δx = had / lim δx → 0 δx → 0 x2 + 2xδx + (δx)2 + x + δx – x2 – x δx = had / lim δx → 0 δx → 0 2xδx + (δx)2 + δx δx = had / lim δx → 0 δx → 0 (2x + δx + 1) = 2x + (0) + 1 = 2x + 1 Latihan 5 Cari dy dx bagi setiap yang berikut dengan menggunakan prinsip pertama. TP 2 Find dy dx for each of the following by using the fi rst principle. TP 2 Mempamerkan kefahaman tentang pembezaan. 1 y = 5x y = 5x }}} 1 y + δy = 5(x + δx) y + δy = 5x + 5δx }}} 2 2 − 1 , δy = 5δx δy δx = 5 dy dx = had / lim δx → 0 δx → 0 5 = 5 2 y = 4x2 y = 4x2 }}} 1 y + δy = 4(x + δx)2 y + δy = 4[x2 + 2xδx + (δx)2 ] y + δy = 4x2 + 8xδx + 4(δx)2 }}} 2 2 − 1 , δy = 8xδx + 4(δx)2 δy δx = 8x + 4δx dy dx = had / lim δx → 0 δx → 0 8x + 4δx = 8x + 4(0) = 8x 3 y = 3 x y = 3 x }}} 1 y + δy = 3 x + δx }}} 2 2 – 1 , δy = 3 x + δx − 3 x δy = 3x − 3(x + δx) x(x + δx) δy = –3δx x(x + δx) δy δx = –3 x(x + δx) dy dx = had / lim δx → 0 δx → 0 – 3 x(x + δx) = – 3 x(x + 0) = – 3 x2 Contoh 5 y = 3x2 Penyelesaian y = 3x2 }}} 1 y + δy = 3(x + δx)2 y + δy = 3[x2 + 2xδx + (δx)2 ] y + δy = 3x2 + 6xδx + 3(δx)2 }}} 2 2 − 1 , δy = 6xδx + 3(δx)2 δy δx = 6x + 3δx dy dx = had /lim δx → 0 δx → 0 6x + 3δx = 6x + 3(0) = 6x dy dx = had / lim δx → 0 δx → 0 δy δx dy dx = had / lim δx → 0 δx → 0 δy δx Tip Bestari Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 22 26/10/2023 12:12:10 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

23 2.2 Pembezaan Peringkat Pertama/ The First Derivative Latihan 6 Cari dy dx atau f'(x) bagi setiap yang berikut. TP 2 Find dy dx or f'(x) for each of the following. TP 2 Mempamerkan kefahaman tentang pembezaan. 1 y = 3x6 dy dx = 6(3x6 – 1) = 18x5 2 y = –6x4 dy dx = 4(–6x4 – 1) = –24x3 3 y = 2 x3 y = 2x–3 dy dx = –3(2x–3 – 1) = –6x–4 = – 6 x4 4 f(x) = x6 2 f ′(x) = 6 1 x6 – 1 2 2 = 3x5 5 f(x) = – 4 x2 f(x) = –4x–2 f′(x) = –2(–4x–2 – 1) = 8x–3 = 8 x3 6 y = 1 3x2 y = x–2 3 dy dx = (–2) 1 x–2 – 1 3 2 = –2x–3 3 = – 2 3x3 7 f(x) = 18x–2 x f(x) = 18x–2 – 1 = 18x–3 f′(x) = (–3)(18x–3 – 1) = –54x–4 = – 54 x4 (a) y = 4x3 (b) f(x) = 3x5 (c) y = 1 x2 Penyelesaian (a) y = 4x3 dy dx = 3(4x3 − 1 ) = 12x2 (b) f(x) = 3x5 f , (x) = 5(3x5 − 1 ) = 15x4 (c) 1 x2 = x–2 dy dx = –2x–2 – 1 = –2x–3 = – 2 x3 Contoh 6 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 23 26/10/2023 12:12:11 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

24 Latihan 7 Tentukan nilai dy dx atau ƒ'(x) bagi setiap yang berikut apabila nilai x diberi. TP 2 Determine the value of dy dx or ƒ'(x) for each of the following when the value of x is given. TP 2 Mempamerkan kefahaman tentang pembezaan. 1 y = 2x4 , x = –2 dy dx = 4(2x4 – 1) = 8x3 x = –2, dy dx = 8(–2)3 = –64 2 y = 2x7 , x = 1 dy dx = 7(2x7 – 1) = 14x6 x = 1, dy dx = 14(1) 6 = 14 3 f(x) = 3x2 , x = 5 f′(x) = 2(3x2 – 1) = 6x f′(5) = 6(5) = 30 4 f(x) = – x6 3 , x = 2 f ′(x) = –61 x6 – 1 3 2 = –2x5 f ′(2) = –2(2)5 = –64 5 f(x) = 3 x2 , x = –1 f(x) = 3x–2 f′(x) = –2(3x–2 – 1) = –6x–3 = – 6 x3 f′(–1) = – 6 (–1)3 = – 6 (–1) = 6 6 f(x) = 9x2 , x = –4 f′(x) = 2(9x2 – 1) = 18x f′(–4) = 18(–4) = –72 7 y = 6 x3 , x = 3 y = 6x–3 dy dx = –3(6x–3 – 1) = –18x–4 = –18 x4 dy dx = –18 (3)4 = – 2 9 8 y = 1 5x7 , x = –2 y = x–7 5 dy dx = –7(x–7 – 1) 5 = –7x–8 5 = –7 5x8 dy dx = –7 5(–2)8 = – 7 1 280 (a) y = x3 6 , x = 4 Penyelesaian (a) y = x3 6 dy dx = 3 1 x3 – 1 6 2 = x2 2 = (4)2 2 = 8 (b) f(x) = 3x4 , x = −2 (b) f(x) = 3x4 f ′(x) = 4(3x4 − 1 ) = 12x3 f ′(−2) = 12(−2)3 = −96 Contoh 7 Sudut Kalkulator Menentukan penyelesaian bagi Contoh 7(b) menggunakan kalkulator saintifi k. Determine the solution of Example 7(b) by using scientifi c calculator. Langkah 1/Step 1: Tekan ‘MENU’ dan cari 1 : Calculate (paparan normal), tekan ‘ = ’. Press ‘MENU’ and fi nd 1 : Calculate (normal display), press ‘ = ’. Langkah 2/Step 2: Tekan/Press: SHIFT ∫ 3 x x 4 (–) 2 = Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 24 26/10/2023 12:12:12 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

25 Latihan 8 Bezakan setiap yang berikut terhadap x. TP 2 Differentiate each of the following with respect to x. TP 2 Mempamerkan kefahaman tentang pembezaan. 1 2x7 − 8x + 9 d dx (2x7 − 8x + 9) = 7(2x7 – 1) – 1(8x1 – 1) = 14x6 − 8 2 4x2 + 3 4x − 5x + 1 9 d dx 14x2 + 3 4x − 5x + 1 92 = 2(4x2 – 1) + (–1)1 3x–1 – 1 4 2 − 5x1 – 1 = 8x − 3x–2 4 − 5 = 8x − 3 4x2 − 5 3 x5 + 2 x d dx (x5 + 2x–1) = 5(x5 – 1) + (–1)(2x–1 – 1) = 5x4 − 2x–2 = 5x4 − 2 x2 4 x7 8 − 2x 3 + 9 d dx 1 x7 8 − 2x 3 + 92 = 71 x7 – 1 8 2 − 2x1 – 1 3 = 7x6 8 − 2 3 5 x4 − 3 x d dx 1 x4 − 3 x 2 = d dx (x3 − 3x–1) = 3(x3 – 1) – (–1)(3x–1 – 1) = 3x2 + 3x–2 = 3x2 + 3 x2 Latihan 9 Cari fungsi kecerunan bagi setiap yang berikut. TP 3 Find the gradient function for each of the following. TP 3 Mengaplikasikan kefahaman tentang pembezaan untuk melaksanakan tugasan mudah. 1 y = x4 − 3x2 + 7x − 12 y = x4 − 3x2 + 7x − 12 dy dx = 4x3 − 6x + 7 2 y = 2x(4 − x2 ) y = 2x(4 − x2 ) y = 8x − 2x3 dy dx = 8 − 6x2 3x4 + 5x − 2 Penyelesaian d dx (3x4 + 5x − 2) = 4(3x4 – 1) + (1)(5x1 – 1 ) = 12x3 + 5 Contoh 8 y = (2x − 1)2 Penyelesaian y = (2x − 1)2 = (2x − 1)(2x − 1) Kembangkan Expand = 4x2 − 4x + 1 dy dx = 8x − 4 dy dx juga dikenali sebagai fungsi kecerunan. dy dx is also known as the gradient function. Tip Bestari Contoh 9 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 25 26/10/2023 12:12:12 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

26 3 y = (3x + 10)2 y = (3x + 10)2 = (3x + 10)(3x + 10) = 9x2 + 60x + 100 dy dx = 18x + 60 4 y = 2(2x + 9)2 3 y = 2 3 (4x2 + 36x + 81) = 8 3 x2 + 24x + 54 dy dx = 16 3 x + 24 5 y = 13x − 5 x2 2 y = 13x − 5 x2 2 = 13x − 5 x213x − 5 x2 = 9x2 − 3x1 5 x2 − 3x1 5 x2 + 25 x2 = 9x2 − 30 + 25x– 2 dy dx = 18x − 50x–3 = 18x − 50 x3 6 y = 1x − 3 x2 2 y = 1x − 3 x2 2 = 1x − 3 x21x − 3 x2 = x2 − 6 + 9 x2 = x2 − 6 + 9x–2 dy dx = 2x − 18x–3 = 2x − 18 x3 7 y = 5x4 + 6 x2 y = 5x4 + 6 x2 = 5x2 + 6x–2 dy dx = 10x − 12x–3 = 10x − 12 x3 8 y = 3x4 − 5x x3 + 2x − 1 y = 3x4 − 5x x3 + 2x − 1 = 3x − 5x–2 + 2x − 1 = 5x − 5x–2 − 1 dy dx = 5 + 10x–3 = 5 + 10 x3 Latihan 10 Bezakan setiap yang berikut terhadap x. TP 2 Differentiate each of the following with respect to x. TP 2 Mempamerkan kefahaman tentang pembezaan. 1 y = (3x − 1)4 y = (3x − 1)4 dy dx = 4(3x − 1)3 d dx (3x − 1) = 4(3x − 1)3 (3) = 12(3x − 1)3 y = (x2 + 4x) 3 Penyelesaian dy dx = 3(x2 + 4x) 2 d dx (x2 + 4x) = 3(x2 + 4x) 2 (2x + 4) = 3(x2 + 4x) 2 • 2(x + 2) = 6(x2 + 4x) 2 (x + 2) Kaedah Alternatif Katakan/ Let u = x2 + 4x ∴ du dx = 2x + 4 y = u3 ∴ dy du = 3u2 dy dx = dy du × du dx = 3u2 × (2x + 4) = 3(x2 + 4x) 2 • 2(x + 2) = 6(x2 + 4x) 2 (x + 2) Contoh 10 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 26 26/10/2023 12:12:13 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

27 y = (2x + 1)(3x2 − 2) Penyelesaian u = 2x + 1 , v = 3x2 − 2 du dx = 2 dv dx = 6x dy dx = u dv dx + v du dx = (2x + 1)(6x) + (3x2 − 2)(2) = 12x2 + 6x + 6x2 − 4 = 18x2 + 6x − 4 Contoh 11 2 y = (3 − 2x) 7 y = (3 − 2x) 7 dy dx = 7(3 − 2x) 6 d dx (3 − 2x) = 7(3 − 2x) 6 (–2) = –14(3 − 2x) 6 3 y = (4x2 + 7)9 y = (4x2 + 7)9 dy dx = 9(4x2 + 7)8 d dx (4x2 + 7) = 9(4x2 + 7)8 (8x) = 72x(4x2 + 7)8 4 y = 3 2x+ 1 y = 3(2x + 1)–1 dy dx = –1[3(2x + 1)–2] d dx (2x + 1) dy dx = –3(2x + 1)–2(2) = – 6 (2x+ 1)2 5 y = 6 (x − 5)4 y = 6(x − 5)–4 dy dx = –4[6(x − 5)–5] d dx (x − 5) = –24(x − 5)–5 ( 1) = – 24 (x − 5)5 Latihan 11 Bezakan setiap yang berikut dengan menggunakan petua hasil darab. TP 2 Differentiate each of the following by using the product rule. TP 2 Mempamerkan kefahaman tentang pembezaan. 1 y = 3x(4x2 – 1) u = 3x , v = 4x2 − 1 du dx = 3 dv dx = 8x dy dx = u dv dx + v du dx = (3x)(8x) + (4x2 − 1)(3) = 24x2 + 12x2 − 3 = 36x2 − 3 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 27 26/10/2023 12:12:14 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

28 2 y = 2x4 (6x2 + 7) u = 2x4 , v = 6x2 + 7 du dx = 8x3 dv dx = 12x dy dx = u dv dx + v du dx = (2x4 )(12x) + (6x2 + 7)(8x3 ) = 24x5 + 48x5 + 56x3 = 72x5 + 56x3 3 y = (2x + 1)(x2 + 4x – 3) u = 2x + 1 , v = x2 + 4x − 3 du dx = 2 dv dx = 2x + 4 dy dx = u dv dx + v du dx = (2x + 1)(2x + 4) + (x2 + 4x − 3)(2) = 4x2 + 8x + 2x + 4 + 2x2 + 8x − 6 = 6x2 + 18x − 2 4 y = (4x − 5)(3x2 + 8) u = 4x − 5 , v = 3x2 + 8 du dx = 4 dv dx = 6x dy dx = u dv dx + v du dx = (4x − 5)(6x) + (3x2 + 8)(4) = 24x2 − 30x + 12x2 + 32 = 36x2 − 30x + 32 5 y = (x – 1)(x2 – 2x + 5) u = x − 1 , v = x2 − 2x + 5 du dx = 1 dv dx = 2x − 2 dy dx = u dv dx + v du dx = (x − 1)(2x − 2) + (x2 − 2x + 5)(1) = 2x2 − 2x − 2x + 2 + x2 − 2x + 5 = 3x2 − 6x + 7 Latihan 12 Bezakan setiap yang berikut dengan menggunakan petua hasil bahagi. TP 2 Differentiate each of the following by using the quotient rule. TP 2 Mempamerkan kefahaman tentang pembezaan. 1 y = 6x 3x+ 2 u = 6x , v = 3x + 2 du dx = 6 dv dx = 3 dy dx = v du dx – u dv dx v2 = (3x + 2)(6) – (6x)(3) (3x + 2)2 = 18x + 12 – 18x (3x + 2)2 = 12 (3x + 2)2 y = 3x − 4 2x2 − 1 Penyelesaian u = 3x − 4 , v = 2x2 − 1 du dx = 3 dv dx = 4x dy dx = v du dx − u dv dx v2 = (2x2 − 1)(3) − (3x − 4)(4x) (2x2 – 1)2 = 6x2 − 3 − 12x2 + 16x (2x2 − 1)2 = 16x − 6x2 − 3 (2x2 − 1)2 Contoh 12 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 28 26/10/2023 12:12:15 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

29 2.3 Pembezaan Peringkat Kedua/ The Second Derivative Latihan 13 Cari d2 y dx2 bagi setiap yang berikut. TP 2 Find d2 y dx2 for each of the following. TP 2 Mempamerkan kefahaman tentang pembezaan. 1 y = x2 (4x − 8) y = x2 (4x − 8) = 4x3 − 8x2 dy dx = 12x2 − 16x d2 y dx2 = 24x − 16 2 y = 7 2x − 1 u = 7 , v = 2x − 1 du dx = 0 dv dx = 2 dy dx = v du dx − u dv dx v2 = (2x − 1)(0) − (7)(2) (2x − 1)2 = – 14 (2x − 1)2 3 y = 3x − 1 4x − 3 u = 3x − 1 , v = 4x − 3 du dx = 3 dv dx = 4 dy dx = v du dx − u dv dx v2 = (4x − 3)(3) − (3x − 1)(4) (4x − 3)2 = 12x − 9 − 12x + 4 (4x − 3)2 = – 5 (4x − 3)2 4 y = 5 − 2x 6x − 1 u = 5 − 2x , v = 6x − 1 du dx = –2 dv dx = 6 dy dx = v du dx − u dv dx v2 = (6x − 1)(–2) − (5 − 2x)(6) (6x – 1)2 = –12x + 2 − 30 + 12x (6x − 1)2 = – 28 (6x – 1)2 5 y = x2 − x − 1 2x+ 3 u = x2 − x − 1 , v = 2x + 3 du dx = 2x − 1 dv dx = 2 dy dx = v du dx − u dv dx v2 = (2x + 3)(2x − 1) − (x2 − x − 1)(2) (2x + 3)2 = 4x2 − 2x + 6x − 3 − 2x2 + 2x + 2 (2x + 3)2 = 2x2 + 6x − 1 (2x + 3)2 y = 7x2 − 2x3 + 5x + 9 Penyelesaian y = 7x2 − 2x3 + 5x + 9 dy dx = 14x − 6x2 + 5 d2 y dx2 = 14 − 12x Contoh 13 d2 y dx2 adalah sama dengan d dx 1 dy dx2. d2 y dx2 is equal to d dx 1 dy dx2. Tip Bestari Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 29 26/10/2023 12:12:15 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

30 2 y = (x2 − 5)2 y = (x2 − 5)2 = x4 − 10x2 + 25 dy dx = 4x3 − 20x d2 y dx2 = 12x2 − 20 3 y = x4 – 3 x y = x4 − 3 x = x4 − 3x–1 dy dx = 4x3 + 3x–2 d2 y dx2 = 12x2 − 6x–3 = 12x2 − 6 x3 Latihan 14 Cari f”(x) bagi setiap yang berikut. TP 2 Find f ”(x) for each of the following. TP 2 Mempamerkan kefahaman tentang pembezaan. 1 f(x) = x2 + 1 x f(x) = x2 + 1 x f ′(x) = 2x − 1 x2 f ″(x) = 2 + 2 x3 2 f(x) = (3x + 1)4 f(x) = (3x + 1)4 f′(x) = 4(3x + 1)3 (3) = 12(3x + 1)3 f ″(x) = 3[12(3x + 1)2 ](3) = 108(3x + 1)2 3 f(x) = 3 (2x − 7)2 f(x) = 3(2x − 7)–2 f′(x) = –2[3(2x − 7)–3](2) = –12(2x − 7)–3 f ″(x) = –3[–12(2x − 7)–4](2) = 72 (2x − 7)4 f(x) = 4 (3x − 5)2 Penyelesaian f(x) = 4 (3x − 5)2 = 4(3x − 5)–2 f ′(x) = –2[4(3x − 5)–3](3) = –24(3x − 5)–3 f ″(x) = –3[–24(3x – 5)–4](3) = 216 (3x − 5)4 f ″ (x) adalah sama dengan d dx [f′(x)]. f “(x) is equal to d dx [f ′(x)]. Tip Bestari Contoh 14 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 30 26/10/2023 12:12:16 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

31 2.4 Aplikasi Pembezaan/ Application of Differentiation Latihan 15 Hitung kecerunan tangen kepada lengkung pada titik-titik yang diberi. TP 3 Calculate the gradient of the tangent to the curve at the points given. TP 3 Mengaplikasikan kefahaman tentang pembezaan untuk melaksanakan tugasan mudah. 1 Given y = x2 + 5. Diberi y = x2 + 5. Titik Point (–3, 14) (–2, 9) (–1, 6) (0, 5) (1, 6) (2, 9) (3, 14) Kecerunan Gradient –6 –4 –2 0 2 4 6 Diberi/Given y = x2 + 5, dy dx = 2x Pada/At (–3, 14), Kecerunan/ Gradient = dy dx = 2(–3) = –6 Pada/At (–2, 9), Kecerunan/ Gradient = dy dx = 2(–2) = –4 Pada/At (–1, 6), Kecerunan/ Gradient = dy dx = 2(–1) = –2 Pada/At (0, 5), Kecerunan/ Gradient = dy dx = 2(0) = 0 Pada/At (1, 6), Kecerunan/ Gradient = dy dx = 2(1) = 2 Pada/At (2, 9), Kecerunan/ Gradient = dy dx = 2(2) = 4 Pada/At (3, 14), Kecerunan/ Gradient = dy dx = 2(3) = 6 Diberi y = –x2 + 6. Given y = –x2 + 6. Titik Point (–3, –3) (–2, 2) (–1, 5) (0, 6) Kecerunan Gradient Penyelesaian Diberi/Given y = –x2 + 6, dy dx = –2x Pada/At (–3, –3), Kecerunan/Gradient = dy dx = –2(–3) = 6 Pada/At (–2, 2), Kecerunan/Gradient = dy dx = –2(–2) = 4 Pada/At (–1, 5), Kecerunan/Gradient = dy dx = –2(–1) = 2 Pada/At (0, 6), Kecerunan/Gradient = dy dx = –2(0) = 0 Titik Point (–3, –3) (–2, 2) (–1, 5) (0, 6) Kecerunan Gradient 6 4 2 0 Contoh 15 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 31 26/10/2023 12:12:17 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

32 Latihan 16 Cari kecerunan tangen, m1 , dan kecerunan normal, m2 , pada titik yang diberi. Seterusnya, cari persamaan tangen dan normal kepada lengkung yang berikut. TP 3 Find the gradient of the tangent, m1 , and the gradient of the normal, m2 , at the given points. Then, fi nd the equation of the tangent and normal to the following curves. TP 3 Mengaplikasikan kefahaman tentang pembezaan untuk melaksanakan tugasan mudah. Kecerunan Gradient Persamaan tangen Equation of the tangent Persamaan normal Equation of the normal 1 y = 3x2 − 4x + 5; (1, 4) dy dx = 6x − 4 = 6(1) − 4 = 2 ∴ m1 = 2, m2 = – 1 2 m1 = 2 y − 4 = 2(x − 1) y − 4 = 2x − 2 y = 2x + 2 m2 = – 1 2 y − 4 = – 1 2 (x − 1) 2(y − 4) = –(x − 1) 2y − 8 = –x + 1 2y = –x + 9 2 y = x2 − 2x + 5; (–4, 3) dy dx = 2x − 2 = 2(–4) − 2 = –10 ∴ m1 = –10, m2 = 1 10 m1 = –10 y − 3 = –10[x − (–4)] y − 3 = –10x − 40 y = –10x − 37 m2 = 1 10 y − 3 = 1 10 [x − (–4)] 10(y − 3) = x + 4 10y − 30 = x + 4 10y = x + 34 3 y = x3 + x − 4; (1, –2) dy dx = 3x2 + 1 = 3(1)2 + 1 = 4 ∴ m1 = 4, m2 = – 1 4 m1 = 4 y − (–2) = 4(x − 1) y + 2 = 4x − 4 y = 4x − 6 m2 = – 1 4 y – (–2) = – 1 4 (x − 1) 4(y + 2) = –(x − 1) 4y + 8 = –x + 1 4y = –x − 7 Kecerunan Gradient Persamaan tangen Equation of the tangent Persamaan normal Equation of the normal y = 2x2 − 5x + 6; (2, 4) Penyelesaian dy dx = 4x − 5 = 4(2) − 5 = 3 ∴ m1 = 3, m2 = – 1 3 m1 = 3 y − y1 = m1 (x − x1 ) y − 4 = 3(x − 2) y − 4 = 3x − 6 y = 3x − 2 m2 = − 1 3 y − y1 = m2 (x − x1 ) y − 4 = − 1 3 (x − 2) 3(y − 4) = –(x − 2) 3y − 12 = –x + 2 3y = –x + 14 Contoh 16 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 32 26/10/2023 12:12:17 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

33 Kecerunan Gradient Persamaan tangen Equation of the tangent Persamaan normal Equation of the normal 4 y = 12 x ; (2, 3) dy dx = – 12 x2 = – 12 (2)2 = –3 ∴ m1 = –3, m2 = 1 3 m1 = –3 y − 3 = –3(x − 2) y − 3 = –3x + 6 y = –3x + 9 m2 = 1 3 y − 3 = 1 3 (x − 2) 3(y − 3) = (x − 2) 3y − 9 = x − 2 3y = x + 7 5 y = (2x − 3)4 ; (2, 1) dy dx = 4(2x − 3)3 (2) = 8(2x − 3)3 = 8[2(2) − 3]3 = 8 ∴ m1 = 8, m2 = – 1 8 m1 = 8 y − 1 = 8(x − 2) y − 1 = 8x − 16 y = 8x − 15 m2 = – 1 8 y − 1 = – 1 8 (x − 2) 8(y − 1) = –(x − 2) 8y – 8 = –x + 2 8y = –x + 10 Latihan 17 Selesaikan masalah yang berikut. TP 3 Solve the following problems. TP 3 Mengaplikasikan kefahaman tentang pembezaan untuk melaksanakan tugasan mudah. Diberi persamaan lengkung ialah y = 7x – x2 , cari Given that the equation of a curve is y = 7x – x2 , fi nd (a) dy dx pada titik P(2, 10),/ dy dx at point P(2, 10), (b) persamaan tangen kepada lengkung pada titik P, the equation of the tangent to the curve at point P, (c) persamaan normal kepada lengkung pada titik P. the equation of the normal to the curve at point P. Penyelesaian (a) y = 7x – x2 dy dx = 7 – 2x Apabila/When x = 2, dy dx = 7 – 2(2) = 3 (b) Kecerunan tangen/ Gradient of the tangent = 3 Persamaan tangen/ Equation of the tangent: y – 10 = 3(x – 2) y – 10 = 3x – 6 y = 3x + 4 (c) Kecerunan normal/ Gradient of the normal = – 1 3 Persamaan normal/Equation of the normal: y – 10 = – 1 3 (x – 2) 3y – 30 = –x + 2 3y = –x + 32 Contoh 17 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 33 26/10/2023 12:12:17 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

34 1 Diberi persamaan lengkung ialah y = 3 2 x2 – 22, cari Given that the equation of a curve is y = 3 2 x2 – 2, find (a) dy dx pada titik P(4, 2),/ dy dx at point P(4, 2), (b) persamaan tangen kepada lengkung pada titik P, the equation of the tangent to the curve at point P, (c) persamaan normal kepada lengkung pada titik P. the equation of the normal to the curve at point P. (a) y = 3 2 x2 – 22 dy dx = 3x Apabila/When x = 4, dy dx = 3(4) = 12 (b) Kecerunan tangen = 12 Gradient of the tangent = 12 Persamaan tangen, Equation of the tangent, y – 2 = 12(x – 4) y – 2 = 12x – 48 y = 12x – 46 (c) Kecerunan normal = – 1 12 Gradient of the normal = – 1 12 Persamaan normal, Equation of the normal, y – 2 = – 1 12 (x – 4) 12y – 24 = –x + 4 12y = –x + 28 2 Diberi persamaan lengkung ialah y = (2x – 3)3 , cari Given that the equation of a curve is y = (2x – 3)3 , find (a) dy dx pada titik P(2, 1),/ dy dx at point P(2, 1), (b) persamaan tangen kepada lengkung pada titik P, the equation of the tangent to the curve at point P, (c) persamaan normal kepada lengkung pada titik P. the equation of the normal to the curve at point P. (a) y = (2x – 3)3 dy dx = 3(2x – 3)2 (2) = 6(2x – 3)2 Apabila/When x = 2, dy dx = 6[2(2) – 3]2 = 6 (b) Kecerunan tangen = 6 Gradient of the tangent = 6 Persamaan tangen, Equation of the tangent, y – 1 = 6(x – 2) y – 1 = 6x – 12 y = 6x – 11 (c) Kecerunan normal = – 1 6 Gradient of the normal = – 1 6 Persamaan normal, Equation of the normal, y – 1 = – 1 6 (x – 2) 6y – 6 = –x + 2 6y = –x + 8 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 34 26/10/2023 12:12:18 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

35 1 Guru menyediakan beberapa set soalan melibatkan tangen dan normal kepada lengkung pada kad berwarna. Teacher provides a set of questions involving tangent and normal to the curve on coloured cards. 2 Murid melakukan aktiviti ini secara berkumpulan yang terdiri daripada tiga orang murid. Sekeping kad berwarna dipilih secara rawak oleh setiap kumpulan. Students perform this activity in groups of three students. A coloured card is randomly selected by each group. 3 Setiap kumpulan dikehendaki menjawab semua soalan yang terdapat pada kad yang dipilih. Tulis jawapan pada kertas mahjung. Each group is required to answer all questions on the selected card. Write the answers on a mahjung paper. 4 Murid menampal hasil kerja pada papan kenyataan. Seorang ketua dalam setiap kumpulan dikehendaki berdiri di sebelah hasil kerja masing-masing. Students paste their work on the notice board. A leader in each group is required to stand next to their group work. 5 Satu kumpulan dikehendaki bergerak ke setiap kumpulan bagi menilai hasil kerja kumpulan yang lain. Setelah selesai, kumpulan lain perlu melakukan langkah yang sama. A group is required to move to each group to evaluate the work of other groups. Once completed, other groups need to do the same. 6 Guru mengadakan perbincangan dengan murid untuk menambahkan kefahaman mereka tentang penyelesaian masalah melibatkan tangen dan normal. Teacher holds a discussion with students to enhance their understanding about solving problems involving tangent and normal. 3 Diberi persamaan lengkung ialah y = 6 x + 8, cari Given that the equation of a curve is y = 6 x + 8, find (a) dy dx pada titik P(–2, 5),/ dy dx at point P(–2, 5), (b) persamaan tangen kepada lengkung pada titik P, the equation of the tangent to the curve at point P, (c) persamaan normal kepada lengkung pada titik P. the equation of the normal to the curve at point P. (a) y = 6 x + 8 dy dx = – 6 x2 Apabila/When x = –2, dy dx = – 6 (–2)2 = – 6 4 = – 3 2 (b) Kecerunan tangen = – 3 2 Gradient of the tangent = – 3 2 Persamaan tangen, Equation of the tangent, y – 5 = – 3 2 [x – (–2)] 2(y – 5) = –3(x + 2) 2y – 10 = –3x – 6 2y = –3x + 4 (c) Kecerunan normal Gradient of the normal, = – 1 (– 3 2 ) = 2 3 Persamaan normal, Equation of the normal, y – 5 = 2 3 [x – (–2)] 3(y – 5) = 2(x + 2) 3y – 15 = 2x + 4 3y = 2x + 19 Aktiviti PAK-21 Gallery Walk PdPc Aktiviti PAK-21 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 35 26/10/2023 12:12:18 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

36 Latihan 18 Tentukan titik pusingan bagi setiap yang berikut. TP 3 Determine the turning point for each of the following. TP 3 Mengaplikasikan kefahaman tentang pembezaan untuk melaksanakan tugasan mudah. y = 2x3 − 6x2 − 18x + 9 Penyelesaian y = 2x3 − 6x2 − 18x + 9 dy dx = 6x2 − 12x − 18 Pada titik pusingan, At turning points, dy dx = 0 6x2 − 12x − 18 = 0 x2 − 2x − 3 = 0 (x + 1)(x – 3) = 0 x = –1 atau/or x = 3 Apabila/ When x = –1, y = 2(–1)3 − 6(–1)2 − 18(–1) + 9 = 19 Apabila/ When x = 3, y = 2(3)3 − 6(3)2 − 18(3) + 9 = –45 Menggunakan pembezaan peringkat kedua, Using the second order derivative, d2 y dx2 = 12x − 12 Apabila/ When x = –1, d2 y dx2 = 12(–1) − 12 = –24 (< 0) y mempunyai nilai maksimum apabila x = –1. y has a maximum value when x = –1. Apabila/ When x = 3, d2 y dx2 = 12(3) − 12 = 24 (> 0) y mempunyai nilai minimum apabila x = 3. y has a minimum value when x = 3. Titik maksimum/Maximum point: (–1, 19) Titik minimum/Minimum point: (3, –45) Contoh 18 Apabila/ When x = –1, y = 2(–1)3 – 6(–1)2 – 18(–1) + 9 = 19 Apabila/ When x = 3, y = 2(3)3 – 6(3)2 – 18(3) + 9 = –45 Penjadualan dan lakaran lengkung Tabulating and curve sketching dy dx = 6x2 – 12x – 18 Apabila/ When = –1, x = –2 (x < –1) x = –1 x = 0 (x > –1) y = 6(–2)2 – 12(–2) – 18 = 30 0 y = 6(0)2 – 12(0) – 18 = –18 + 0 – Lengkung/ Curve: Titik maksimum/ Maximum point: (–1, 19) Apabila/ When x = 3, x = 2 (x < 3) x = 3 x = 4 (x > 3) y = 6(2)2 – 12(2) – 18 = –18 0 y = (4)2 – 12(4) – 18 = 30 – 0 + Lengkung/ Curve: Titik minimum/ Minimum point: (3, −45) Kaedah Alternatif Untuk tujuan pembelajaran Imbas kod QR atau layari htt ps://bestmaths.net/ online/index.php/year-levels/year-12/year-12- topic-list/stati onary-and-turning-points/ untuk nota tambahan tentang � � k pegun dan � � k pusingan. Laman Web Laman Web 1 y = 6x − x2 + 7 y = 6x − x2 + 7 dy dx = 6 − 2x Pada titik pusingan, / At turning points, dy dx = 0 6 − 2x = 0 –2x = –6 x = 3 d2 y dx2 = –2 (< 0) y mempunyai nilai maksimum apabila x = 3. y has a maximum value when x = 3. y = 6(3) − (3)2 + 7 = 16 Titik pusingan ialah titik maksimum (3, 16). The turning point is a maximum point (3, 16). Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 36 26/10/2023 12:12:19 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

37 4 y = 5 – 1 x − 4x y = 5 − x–1 − 4x dy dx = x–2 − 4 = 1 x2 − 4 Pada titik pusingan, At turning points, dy dx = 0 1 x2 − 4 = 0 1 − 4x2 = 0 4x2 − 1 = 0 (2x − 1)(2x + 1) = 0 x = 1 2 atau/or x = – 1 2 d2 y dx2 = –2x–3 = –2 x3 Apabila/When x = 1 2, d2 y dx2 = –2 1 1 22 3 = –16 (< 0) y mempunyai nilai maksimum apabila x = 1 2. y has a maximum value when x = 1 2. y = 5 − 1 1 1 22 − 4 1 1 22 = 1 Apabila/When x = – 1 2, d2 y dx2 = –2 1– 1 22 3 = 16 (> 0) y mempunyai nilai minimum apabila x = – 1 2 . y has a minimum value when x = – 1 2. y = 5 − 1 1– 1 22 – 4 1– 1 22 = 9 Titik maksimum: 1 1 2, 12 Maximum point: 1 1 2 , 12 Titik minimum: 1– 1 2, 92 Minimum point:1– 1 2 , 92 2 y = 1 3 x3 − x2 − 15x + 6 y = 1 3 x3 − x2 − 15x + 6 dy dx = x2 − 2x − 15 Pada titik pusingan, / At turning points, dy dx = 0 x2 − 2x − 15 = 0 (x + 3)(x − 5) = 0 x = –3 atau/or x = 5 d2 y dx2 = 2x − 2 Apabila/ When x = –3, d2 y dx2 = 2(–3) − 2 = –8 (< 0) y mempunyai nilai maksimum apabila x = –3. y has a maximum value when x = –3. y = 1 3(–3)3 − (–3)2 − 15(–3) + 6 = 33 Apabila/ When x = 5, d2 y dx2 = 2(5) − 2 = 8 (> 0) y mempunyai nilai minimum apabila x = 5. y has a minimum value when x = 5. y = 1 3(5)3 − (5)2 − 15(5) + 6 = –52 1 3 Titik maksimum/Maximum point: (–3, 33) Titik minimum/Minimum point: 15, –52 1 3 2 3 y = 2x + 8 x y = 2x + 8 x dy dx = 2 − 8 x2 Pada titik pusingan, / At turning points, dy dx = 0 2 − 8 x2 = 0 2x2 − 8 = 0 x2 − 4 = 0 (x + 2)(x − 2) = 0 x = –2 atau/or x = 2 d2 y dx2 = 16 x3 Apabila/ When x = –2, d2 y dx2 = 16 (–2)3 = –2 (< 0) y mempunyai nilai maksimum apabila x = –2. y has a maximum value when x = –2. y = 2(–2) + 8 (–2) = –8 Apabila/ When x = 2, d2 y dx2 = 16 (2)3 = 2 (> 0) y mempunyai nilai minimum apabila x = 2. y has a minimum value when x = 2. y = 2(2)+ 8 (2) = 8 Titik maksimum/Maximum point: (–2, –8) Titik minimum/Minimum point: (2, 8) Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 37 26/10/2023 12:12:19 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

38 Sebuah silinder tertutup dengan jejari tapak r cm dan tinggi h cm mempunyai isi padu 27 cm3 . Tunjukkan bahawa jumlah luas permukaan, A, ialah A closed cylinder with a base radius of r cm and a height of h cm has a volume of 27 cm3 . Show that the total surface area, A, is A = 54 r + 2πr2 Seterusnya, cari nilai r apabila A adalah minimum. Hence, fi nd the value of r when A is minimum. Penyelesaian Isi padu/Volume = 27 cm3 πr2 h = 27 h = 27 πr2 Jumlah luas permukaan,/Total surface area, A = 2πrh + 2πr2 = (2πr) 1 27 πr2 2 + 2πr2 = 54 r + 2πr2 (Tertunjuk/Shown) A = 54r–1 + 2πr2 dA dr = – 54 r2 + 4πr dA dr = 0 –54 r2 + 4πr = 0 –54 + 4πr3 = 0 r3 = 27 2π r = r = 3 2π 3 d2 A dr2 = 108 r3 + 4π = 108 1 3 2π 3 2 3 + 4π = 108 1 27 2π 2 + 4π = 12π (> 0) A mempunyai nilai minimum apabila r = 3 2π 3 . A has a minimum value when r = 3 2π 3 . Contoh 19 Latihan 19 Selesaikan setiap yang berikut. TP 5 KBAT Menilai Solve each of the following. TP 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pembezaan dalam konteks penyelesaian masalah rutin yang kompleks. 1 Sebuah kuboid mempunyai tapak segi empat sama bersisi x cm dan tinggi y cm. Diberi isi padu kuboid itu ialah 216 cm3 . Tunjukkan bahawa jumlah luas permukaan, A, ialah A cuboid has a square base with sides of x cm and a height of y cm. Given the volume of the cuboid is 216 cm3 . Show that the total surface area, A, is A = 2x2 + 864 x Seterusnya, cari nilai x apabila A adalah minimum. Hence, fi nd the value of x when A is minimum. Isi padu = panjang (x) × lebar (x) × tinggi (y) Volume = length (x) × width (x) × height (y) 216 = x2 y y = 216 x2 Jumlah luas permukaan/ Total surface area A = (x)(x) + (x)(x) + (x)(y) + (x)(y) + (x)(y) + (x)(y) = 2x2 + 4xy = 2x2 + 4x 1 216 x2 2 = 2x2 + 864 x (Tertunjuk /Shown) x y x dA dx = 4x − 864 x2 dA dx = 0 4x − 864 x2 = 0 4x3 − 864 = 0 x3 = 864 4 = 216 x = 216 3 x = 6 d2 A dx2 = 4 + 1 728 x3 = 4 + 1 728 (6)3 = 12 (> 0) A mempunyai luas minimum apabila x = 6. A has a minimum value when x = 6. 27 2π �3 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 38 26/10/2023 12:12:20 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

39 2 Rajah di bawah menunjukkan sebuah silinder tertutup dengan jejari r cm dan tinggi h cm. Isi padu silinder itu ialah 64 cm3 . The diagram below shows a closed cylinder with a radius of r cm and a height of h cm. The volume of the cylinder is 64 cm3 . r cm h cm • (a) Tunjukkan bahawa jumlah luas permukaan, A, ialah Show that the total surface area, A, is A = 128 r + 2πr2 (b) Seterusnya, cari nilai r apabila A adalah minimum. Hence, find the value of r when A is minimum. (a) Isi padu/Volume, πr2 h = 64 h = 64 πr2 Jumlah luas permukaan/ Total surface area, A = 2πrh + 2πr2 = (2πr)1 64 πr2 2 + 2πr2 = 128 r + 2πr2 (Tertunjuk/Shown) (b) dA dr = –128 r2 + 4πr dA dr = 0 –128 r2 + 4πr = 0 –128 + 4πr3 = 0 r3 = 32 π r = 3 32 π d2 A dr2 = 256 r3 + 4π = 256 13 32 π 2 3 + 4π = 256 1 32 π 2 + 4π = 12π (. 0) A mempunyai luas minimum apabila r = 3 32 π . A has a minimum value when r = 3 32 π . 3 Rajah di bawah menunjukkan sebuah bulatan berjejari 12 cm. Sebuah silinder dengan tinggi h cm terterap di dalam bulatan itu. The diagram below shows a circle with a radius of 12 cm. A cylinder with a height of h cm is inscribed in the circle. h cm r cm Cari tinggi silinder apabila isi padunya adalah maksimum. Find the height of the cylinder when its volume is maximum. r cm h cm O 12 cm 1 2 h 1 1 2 h2 2 + r2 = 122 h2 4 + r2 = 144 r2 = 144 − h2 4 Vsilinder/cylinder = πr2 h = πh1144 − h2 4 2 = π1144h − h3 4 2 dV dh = π1144 − 3h2 4 2 dV dh = 0 π1144 − 3h2 4 2 = 0 3h2 4 = 144 h2 = 144 × 4 3 h2 = 192 h = √192 = 8 3 cm d2 V dh2 = π(0 – 3h 2 ) = – 3h 2 π = – 3(8 3 ) 2 π = –12 3 π (, 0) Isi padu adalah maksimum apabila h = 8 3 cm. Volume is maximum when h = 8 3 cm. Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 39 26/10/2023 12:12:21 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

40 Latihan 20 Selesaikan setiap masalah yang berikut. TP 5 KBAT Menilai Solve each of the following problems. TP 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pembezaan dalam konteks penyelesaian masalah rutin yang kompleks. 1 Diberi bahawa y = x3 – 4x dan nilai bagi x berubah pada kadar 0.6 unit s–1. Cari kadar perubahan dalam y apabila x = 3. It is given that y = x3 – 4x and the value of x changes at a rate of 0.6 unit s–1. Find the rate of change in y when x = 3. dx dt = 0.6 y = x3 − 4x dy dx = 3x2 − 4 Apabila/ When x = 3, dy dx = 3(3)2 − 4 = 23 dy dt = dy dx × dx dt = 23 × 0.6 = 13.8 unit s–1/units s–1 2 Jejari sebuah bulatan menokok pada kadar 0.3 cm s–1. Cari kadar perubahan luas bulatan apabila jejari ialah 4 cm. Beri jawapan dalam sebutan π. The radius of a circle increases at a rate of 0.3 cm s–1. Find the rate of change of the area of the circle when the radius is 4 cm. Give the answer in terms of π. dr dt = 0.3 A = πr2 dA dr = 2πr Apabila/ When r = 4, dA dr = 2π(4) = 8π dA dt = dA dr × dr dt = 8π × 0.3 = 2.4π cm2 s–1 3 Rajah di bawah menunjukkan sebuah segi empat sama. The diagram below shows a square. 2x cm Luas segi empat sama itu menyusut pada kadar 3.2 cm2 s–1. Hitung kadar perubahan panjang sisi segi empat sama itu apabila x = 5 cm. The area of the square decreases at a rate of 3.2 cm2 s–1. Calculate the rate of change of the length of the square when x = 5 cm. Luas/Area, A = 2x × 2x = 4x2 dA dx = 8x Apabila/ When x = 5, dA dx = 8(5) = 40 Diberi/Given dA dt = 3.2 dA dt = dA dx × dx dt 3.2 = 40 × dx dt dx dt = 0.08 cm s–1 Isi padu sebuah sfera menyusut pada kadar 5.4π cm3 s–1. Cari kadar perubahan jejari sfera itu apabila jejari ialah 6 cm. The volume of a sphere decreases at a rate of 5.4π cm3 s–1. Find the rate of change in the radius of the sphere when the radius is 6 cm. Penyelesaian dV dt = –5.4π V = 4 3 πr3 dV dr = 4πr2 Apabila/ When r = 6, dV dr = 4π(6)2 = 144π dV dt = dV dr × dr dt –5.4π = 144π × dr dt dr dt = –0.0375 cm s–1 Contoh 20 Kesilapan Umum dV dt = 5.4π Tanda negatif pada −5.4π bermaksud penyusutan isi padu sfera. The negative sign of −5.4π means the decreasing of the volume of the sphere. Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 40 26/10/2023 12:12:22 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

41 Latihan 21 Selesaikan setiap masalah yang berikut. TP 5 KBAT Menilai Solve each of the following problems. TP 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pembezaan dalam konteks penyelesaian masalah rutin yang kompleks. 1 Rajah di bawah menunjukkan air dimasukkan ke dalam sebuah bekas berbentuk hemisfera. The diagram below shows the water is poured into a hemispherical container. 26 cm h cm (a) Ungkapkan luas permukaan air, L dalam sebutan h. Express the area of water surface, L in terms of h. (b) Diberi bahawa kadar perubahan tinggi air ialah 0.5 cm s–1. Cari kadar perubahan luas permukaan air apabila h = 6 cm. It is given that the rate of change of the height of water is 0.5 cm s–1. Find the rate of change of the area of water surface when h = 6 cm. [Guna/ Use π = 3.142] (a) Jejari hemisfera = 13 cm Radius of hemisphere = 13 cm Katakan r = Jejari luas permukaan air Let r = Radius of the water surface 26 cm 13 h cm r 13 - h Dengan menggunakan teorem Pythagoras, By using Pythagoras’ theorem, (13 – h) 2 + r 2 = 132 132 – 26h + h2 + r 2 = 132 r 2 = 26h – h2 L = πr2 = π(26h – h2 ) = 26πh – πh2 (b) dL dh = 26π – 2πh Apabila/When h = 6, dL dh = 26π – 2π(6) = 14π Diberi/Given dh dt = 0.5 dL dt = dL dh × dh dt dL dt = 14π × 0.5 = 7π cm2 s–1 = 21.99 cm2 s–1 Rajah di bawah menunjukkan sebuah bekas berbentuk kon yang berjejari 3 cm dan tinggi 12 cm. The diagram below shows a conical container with a radius of 3 cm and a height of 12 cm. 12 cm 3 cm h cm (a) Ungkapkan isi padu air, V dalam sebutan h. Express the volume of water, V in terms of h. (b) Diberi bahawa air mengalir keluar melalui lubang bekas itu pada kadar tetap 1.8 cm3 s–1. Cari kadar perubahan ketinggian paras air apabila h = 3.6 cm. It is given that the water leaks out from the hole of the container at a constant rate of 1.8 cm3 s–1. Find the rate of change of the height of the water level when h = 3.6 cm. [Guna/ Use π = 3.142] Penyelesaian (b) Katakan r = jejari permukaan air Let r = radius of the water surface r h = 3 12 r = h 4 V = 1 3 × π × r2 × h = 1 3 × π × ( h 4 ) 2 × h = 1 48 πh3 (b) dV dh = 1 16 πh2 Apabila/When h = 3.6, dV dh = 1 16 π(3.6)2 = 0.81 π Diberi/Given dV dt = –1.8 dV dt = dV dh × dh dt –1.8 = 0.81π × dh dt dh dt = –1.8 0.81π = – 20 9π cm s–1 = –0.7073 cm s–1 Contoh 21 1 Jika V menyusut, maka dV dt adalah negatif. If V decreases, then dV dt is negative. 2 Jika V menokok, maka dV dt adalah positif. If V increases, then dV dt is positive. Tip Bestari Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 41 26/10/2023 12:12:23 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

42 2 Rajah di bawah menunjukkan air dimasukkan ke dalam sebuah bekas berbentuk kon yang berjejari 6 cm dan tinggi 18 cm. The diagram below shows the water is poured into a conical container with a radius of 6 cm and a height of 18 cm. 18 cm 6 cm h cm (a) Ungkapkan isi padu air, V dalam sebutan h. Express the volume of water, V in terms of h. (b) Diberi bahawa kadar perubahan ketinggian paras air ialah 0.72 cm s–1. Cari kadar perubahan isi padu air apabila h = 4 cm. It is given that the rate of change of the height of water level is 0.72 cm s–1. Find the rate of change of the volume of water when h = 4 cm. [Guna/ Use π = 3.142] (a) Katakan r = Jejari luas permukaan air Let r = Radius of the water surface r h = 6 18 r = h 3 V = 1 3 × π × r2 × h = 1 3 × π × ( h 3 ) 2 × h = 1 27 πh3 (b) dV dh = 1 9 πh2 Apabila/When h = 4, dV dh = 1 9 π(4)2 = 16 9 π Diberi/Given dh dt = 0.72 dV dt = dV dh × dh dt dV dt = 16 9 π × 0.72 = 32 25 π = 4.022 cm3 s–1 3 Udara daripada sebuah belon berbentuk sfera dilepaskan pada kadar tetap 5π cm3 s–1. Cari kadar perubahan jejari belon apabila jejari ialah 10 cm. [Isi padu sfera = 4 3 πr3 ] The air from a spherical balloon is released at a constant rate of 5π cm3 s–1. Find the rate of change of the radius of the balloon when the radius is 10 cm. [Volume of sphere = 4 3 πr 3 ] Katakan V = Isi padu belon Let V = Volume of the balloon V = 4 3 πr 3 dV dr = 4πr 2 Apabila/When r = 10, dV dr = 4π(10)2 = 400π Diberi/Given dV dt = –5π dV dt = dV dr × dr dt –5π = 400π × dr dt dr dt = – 5π 400π dr dt = –0.0125 cm s–1 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 42 26/10/2023 12:12:25 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

43 Latihan 22 Selesaikan setiap masalah yang berikut. TP 3 Solve each of the following problems. TP 3 Mengaplikasikan kefahaman tentang pembezaan untuk melaksanakan tugasan mudah. 1 Diberi y = x3 , cari nilai dy dx apabila x = 5. Given y = x3 , fi nd the value of dy dx when x = 5. Seterusnya, cari nilai hampir bagi Hence, fi nd the approximate value of (a) 5.053 , (b) 4.953 . y = x3 , dy dx = 3x2 Apabila/When x = 5, y = (5)3 = 125 dy dx = 3(5)2 = 75 (a) dx = 5.05 – 5 = 0.05 x3 = y (x + dx)3 = y + dy (5 + 0.05)3 = (125) + dy dx × dx (5.05)3 = 125 + (75) × 0.05 = 125 + 3.75 = 128.75 (b) dx = 4.95 – 5 = –0.05 x3 = y (x + dx) 3 = y + dy [5 + (–0.05)]3 = 125 + dy dx × dx 4.953 = 125 + (75) × (–0.05) = 125 – 3.75 = 121.25 2 Diberi y = x , cari nilai dy dx apabila x = 16. Given y = x , fi nd the value of dy dx when x = 16. Seterusnya, cari nilai hampir bagi Hence, fi nd the approximate value of (a) 16.2 , (b) 15.8 . y = x 1 2 , dy dx = 1 2 x–1 2 Apabila/When x = 16, y = 161 2 = 4 dy dx = 1 2 (16)–1 2 = 1 2 ( 1 4 ) = 1 8 (a) dx = 16.2 – 16 = 0.2 x = y x + dx = y + dy 16 + 0.2 = (4) + dy dx × dx 16.2 = (4) + ( 1 8 ) × 0.2 = 4.025 (b) dx = 15.8 – 16 = –0.2 x = y x + dx = y + dy 16 + (–0.2) = (4) + dy dx × dx 15.8 = (4) + ( 1 8 ) × (–0.2) = 3.975 Diberi y = x 4 , cari nilai dy dx apabila x = 81. Given y = x 4 , fi nd the value of dy dx when x = 81. Seterusnya, cari nilai hampir bagi Hence, fi nd the approximate value of (a) 81.04 4 , (b) 80.96 4 . Penyelesaian y = x 1 4 dy dx = 1 4 x–3 4 Apabila/When x = 81, y = 81 1 4 = 3 dy dx = 1 4 (81)– 3 4 = 1 4 ( 1 27) = 1 108 (a) 81.04 4 x = 81, dx = 81.04 – 81 = 0.04 x4 = y x + dx 4 = y + dy 81 + 0.04 4 = (3) + dy dx × dx 81.04 4 = (3) + ( 1 108) × 0.04 = (3) + 0.00037 = 3.00037 (b) 80.96 4 x = 81, dx = 80.96 – 81 = –0.04 x4 = y x + dx 4 = y + dy 81 + (–0.04) 4 = (3) + dy dx × dx 80.96 4 = (3) + ( 1 108) × (–0.04) = (3) – 0.00037 = 2.9996 Contoh 22 Tip Bestari δy δx dy dx δy dy dx × δx Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 43 26/10/2023 12:12:26 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

44 3 Diberi y = x 3 , cari nilai dy dx apabila x = 64. Given y = x 3 , fi nd the value of dy dx when x = 64. Seterusnya, cari nilai hampir bagi Hence, fi nd the approximate value of (a) 64.3 3 , (b) 63.7 3 . y = x 1 3 , dy dx = 1 3 x–2 3 Apabila/When x = 64, y = 641 3 = 4 dy dx = 1 3 (64)–2 3 = 1 3 ( 1 16) = 1 48 (a) dx = 64.3 – 64 = 0.3 x 3 = y x + dx 3 = y + dy 64 + 0.3 3 = (4) + dy dx × dx 64.3 3 = (4) + ( 1 48) × 0.3 = (4) + 0.00625 = 4.006 Latihan 23 Selesaikan setiap masalah yang berikut. TP 4 KBAT Mengaplikasi Solve each of the following problems. TP 5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pembezaan dalam konteks penyelesaian masalah rutin yang kompleks. 1 Diberi y = 2x3 + 5x2 + 1, cari perubahan kecil dalam y apabila x menyusut daripada 5 kepada 4.98. Seterusnya, cari nilai hampir bagi y selepas perubahan dalam x berlaku. Given y = 2x3 + 5x2 + 1, fi nd the small change in y when x decreases from 5 to 4.98. Hence, fi nd the approximate value of y after the change in x has taken place. y = 2x3 + 5x2 + 1 dy dx = 6x2 + 10x Apabila/ When x = 5, dy dx = 6(5)2 + 10(5) = 200 δx = 4.98 − 5 = –0.02 δy ≈ dy dx × δx = 200(–0.02) = –4 Nilai hampir bagi y Approximate value of y = [2(5)3 + 5(5)2 + 1] + (–4) = 376 − 4 = 372 (b) dx = 63.7 – 64 = –0.3 x3 = y x + dx 3 = y + dy 64 + (–0.3) 3 = (4) + dy dx × dx 63.7 3 = (4) + ( 1 48) × (–0.3) = (4) – 0.00625 = 3.994 Diberi y = x3 , cari perubahan kecil dalam y apabila x menokok daripada 4 kepada 4.003. Seterusnya, cari nilai hampir bagi y selepas perubahan dalam x berlaku. Given y = x3 , fi nd the small change in y when x increases from 4 to 4.003. Hence, fi nd the approximate value of y after the change in x has taken place. Penyelesaian y = x3 dy dx = 3x2 Apabila/ When x = 4, dy dx = 3(4)2 = 48 δx = 4.003 − 4 = 0.003 δy ≈ dy dx × δx = (48)(0.003) = 0.144 Nilai hampir bagi y Approximate value of y = y + δy = x3 + 0.144 = (4)3 + 0.144 = 64.144 Contoh 23 Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 44 26/10/2023 12:12:27 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

45 2 Diberi y = 6 2x – 1 , cari perubahan kecil dalam y apabila x menokok daripada 2 kepada 2.03. Seterusnya, cari nilai hampir bagi y selepas perubahan dalam x berlaku. Given y = 6 2x − 1 , fi nd the small change in y when x increases from 2 to 2.03. Hence, fi nd the approximate value of y after the change in x has taken place. y = 6 2x − 1 y = 6(2x − 1)–1 dy dx = –6(2x − 1)–2(2) = –12(2x − 1)–2 = – 12 (2x − 1)2 Apabila/When x = 2, dy dx = – 12 [2(2) − 1]2 = – 12 32 = – 4 3 3 Diberi y = – 12 x + 6, cari perubahan kecil dalam x, dalam sebutan k, apabila nilai y menokok daripada 5 kepada 5 + k. Given y = – 12 x + 6, fi nd the small change in x, in terms of k, when the value of y increases from 5 to 5 + k. δy = (5 + k) − 5 = k y = – 12 x + 6 xy = –12 + 6x xy – 6x = –12 x(y – 6) = –12 x = –12 y – 6 x = 12 6 − y x = 12(6 – y) –1 Latihan 24 Selesaikan setiap masalah yang berikut. TP 4 Solve each of the following problems. TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang pembezaan dalam konteks penyelesaian masalah rutin yang mudah. 1 Jejari bagi sebuah bulatan bertambah daripada 8 cm kepada 8.03 cm. Cari luas bulatan yang hampir selepas perubahan dalam jejari berlaku. The radius of a circle increases from 8 cm to 8.03 cm. Find the approximate value of the area of the circle after the change in radius has taken place. A = πr2 dA dr = 2πr Apabila/ When r = 8, dA dr = 2π(8) = 16π δr = 8.03 − 8 = 0.03 δA ≈ dA dr × δr = (16π)(0.03) = 0.48π Nilai hampir bagi A Approximate value of A = π(8)2 + 0.48π = 64.48π cm2 δx = 2.03 − 2 = 0.03 δy ≈ dy dx × δx = 1– 4 32(0.03) = –0.04 Nilai hampir bagi y Approximate value of y = 6 2(2) − 1 + (–0.04) = 2 − 0.04 = 1.96 Cari perubahan kecil dalam isi padu sebuah hemisfera apabila jejarinya berubah daripada 6 cm kepada 6.02 cm. Find the small change in the volume of a hemisphere when its radius changes from 6 cm to 6.02 cm. Penyelesaian V = 2 3 πr3 dV dr = 2πr2 Apabila/When r = 6, dV dr = 2π(6)2 = 72π δr = 6.02 − 6 = 0.02 δV ≈ dV dr × δr = (72π)(0.02) = 1.44π Isi padu hemisfera: Volume of a hemisphere: 1 2 × 4 3 πr3 = 2 3 πr3 Contoh 24 dx dy = –12(6 − y) –2 (–1) = 12(6 − y)–2 = 12 (6 – y)2 Apabila/When y = 5, dx dy = 12 (6 − 5)2 = 12 δx ≈ dx dy × δy = 12(k) = 12k Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 45 26/10/2023 12:12:28 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

46 2 Jejari bagi sebuah sfera berubah daripada 3 cm kepada 2.96 cm. Cari luas permukaan sfera yang hampir selepas perubahan dalam jejari berlaku. The radius of a sphere changes from 3 cm to 2.96 cm. Find the approximate value of the surface area of sphere after the change in radius has taken place. A = 4πr2 dA dr = 8πr Apabila/ When r = 3, dA dr = 8π(3) = 24π δr = 2.96 − 3 = –0.04 δA ≈ dA dr × δr = (24π)(–0.04) = –0.96π Nilai hampir bagi A Approximate value of A = 4πr2 + δA = 4π(3)2 + (–0.96π) = 35.04π cm2 3 Jejari bagi sebuah sfera berubah daripada 3 m kepada 3.06 m. Cari isi padu sfera yang hampir selepas perubahan dalam jejari berlaku. The radius of a sphere changes from 3 m to 3.06 m. Find the approximate value of the volume of the sphere after the change in radius has taken place. V = 4 3 πr3 dV dr = 4πr2 Apabila/ When r = 3, dV dr = 4π(3)2 = 36π δr = 3.06 − 3 = 0.06 δV ≈ dV dr × δr = (36π)(0.06) = 2.16π Nilai hampir bagi V Approximate value of V = V + δV = 4 3 π(3)3 + 2.16π = 38.16π m3 4 Jika jejari sebuah sfera menokok daripada 4.5 cm kepada 4.7 cm, cari perubahan kecil dalam isi padu sfera itu. If the radius of a sphere increases from 4.5 cm to 4.7 cm, fi nd the small change in the volume of the sphere. Isi padu sfera, V Volume of sphere, V V = 4 3 πr3 dV dr = 4πr2 Apabila/When r = 4.5, dV dr = 4π(4.5)2 = 81π δV = dV dr × δr = 81π × (4.7 − 4.5) = 81π × 0.2 = 16.2π cm3 5 Jamal membeli segulung dawai berduri yang panjangnya 68 m. Dia ingin menggunakan kesemua dawai berduri itu untuk memagar sebuah kawasan yang berbentuk segi empat tepat seperti ditunjukkan dalam rajah di bawah untuk menanam ubi keledek. Cari luas maksimum, dalam m2 , kawasan tanaman ubi keledek itu. KBAT Mengaplikasi Jamal bought a roll of barbed wire of length 68 m. He wants to use all the barbed wire to fence up a rectangular area as shown in the diagram below to plant sweet potatoes. Find the maximum area, in m2 , of the sweet potatoes fi eld. Dinding/ Wall Dawai berduri Barbed wire y m x m x + 2y = 68 y = 1 2(68 − x) A = xy = x[ 1 2(68 − x)] = 34x − 1 2 x2 dA dx = 34 − x Apabila/When dA dx = 0, 34 − x = 0 x = 34 y = 1 2[68 − (34)] = 17 A = (34)(17) = 578 m2 d2 A dx2 = –1 < 0 fi Luas maksimum ialah 578 m2 . The maximum area is 578 m2 . Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 46 26/10/2023 12:12:29 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

47 Kertas 1 Bahagian A 1 (a) Nilaikan had x → 3 2x – 6 x2 – x – 6 . Evaluate lim x → 3 2x – 6 x2 – x – 6 . [2 markah/marks] (b) Diberi g(x) = x2 + k 5x – 4 dan g'(1) = 12, cari nilai k. Given that g(x) = x2 + k 5x – 4 and g'(1) = 12, find the value of k. [3 markah/marks] 2 Diberi persamaan bagi suatu lengkung ialah y + 13 = 15x − x3 . It is given that the equation of a curve is y + 13 = 15x − x3 . (a) Cari fungsi kecerunan bagi lengkung itu. Find the gradient function of the curve. [1 markah/mark] (b) Satu garis lurus menyentuh lengkung itu pada titik G dengan keadaan x . 0. Jika garis lurus itu berserenjang dengan garis lurus y + 3x − 2 = 0, cari koordinat titik G. A straight line touches the curve at point G such that x . 0. If the straight line is perpendicular to the straight line y + 3x − 2 = 0, find the coordinates of point G. [3 markah/marks] 3 Diberi x = (3 − 2t²)² dan y = 3t² + 2t, cari Given that x = (3 − 2t²)² and y = 3t² + 2t, find (a) dx dt , (b) dy dx dalam sebutan t. dy dx in terms of t. [4 markah/marks] 4 Encik Afiq hendak memagar sebuah kawasan yang berbentuk segi empat tepat di halaman rumahnya. Kawasan itu berukuran 4x m panjang dan (7 – 2x) m lebar. Cari jumlah panjang, dalam m, pagar yang diperlukan supaya luas kawasan yang dipagar adalah maksimum. Encik Afiq wants to fence up a rectangular area in his yard. The dimensions of the region are 4x m in length and (7 – 2x) m in width. Find the total length, in m, of the fence required in order for the area of the region that will be fenced to be maximum. [4 markah/marks] Bahagian B 5 (a) Cari semua persamaan tangen kepada lengkung y = x x – 2 yang selari dengan garis lurus 3x 2 + 3y = 5. Find all the equations of the tangents to the curve y = x x – 2 that parallel to the straight line 3x 2 + 3y = 5. [5 markah/marks] (b) Rajah 1 menunjukkan lengkung y = 8x − 2x² + 4. Diagram 1 shows a curve y = 8x − 2x² + 4. 0 3 P y x y = 8x – 2x2 + 4 • Rajah 1/ Diagram 1 Cari persamaan normal kepada lengkung itu pada titik P. Find the equation of the normal to the curve at point P. [3 markah/marks] Praktis Berformat SPM Kertas 2 Bahagian A 1 Diberi lengkung y = x + h x , dengan keadaan h ialah pemalar melalui titik (–4, –5). Given the curve y = x + h x , such that h is a constant passes through point (–4, –5). (a) Cari nilai h. Find the value of h. [2 markah/marks] • Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 47 26/10/2023 12:12:29 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA

48 (b) Cari koordinat bagi semua titik pusingan bagi lengkung itu. Find the coordinates of all the turning points of the curve. [4 markah/marks] (c) Lakarkan lengkung itu. Sketch the curve. [2 markah/marks] 2 Rajah 1 menunjukkan sebuah taman berbentuk segi empat tepat. ABFCD ialah halaman rumput, manakala BFC ialah sebuah tasik. Diagram 1 shows a rectangular garden. ABFCD is a lawn, while BFC is a lake. A B D C w meter/ metres 2r meter/ metres FF Rajah 1/ Diagram 1 (a) Diberi panjang pagar yang mengelilingi kawasan halaman rumput ialah 120 m. Tunjukkan bahawa luas, A, bagi kawasan halaman rumput diwakili oleh A = 120r – (2 + 3 2 π)r 2 . Given the length of the fence around the lawn is 120 m. Show that the area, A of the lawn is represented by A = 120r – (2 + 3 2 π)r2 . [2 markah/marks] (b) Cari nilai r dengan keadaan A adalah maksimum. Find the value of r such that A is maximum. [4 markah/marks] (c) Cari luas maksimum, dalam m2 , kawasan halaman rumput itu. Find the maximum area, in m2 , of the lawn. [Guna/Use π = 3.142] [2 markah/marks] 3 Diberi persamaan bagi suatu lengkung ialah y = 21 x3 . Given the equation of a curve is y = 21 x3 . (a) Cari nilai dy dx apabila x = 2. Find the value of dy dx when x = 2. [3 markah/marks] (b) Seterusnya, tentukan nilai hampir bagi 21 (1.98)3 betul kepada dua tempat perpuluhan. Hence, determine the approximate value of 21 (1.98)3 correct to two decimal places. [4 markah/marks] 4 Diberi persamaan bagi suatu lengkung ialah y = 5 + x2 (6 – x). Given the equation of a curve is y = 5 + x2 (6 – x). (a) Tentukan fungsi kecerunan bagi lengkung itu. Determine the gradient function of the curve. [2 markah/marks] (b) Cari koordinat bagi semua titik pusingan. Find the coordinates of all the turning points. [3 markah/marks] (c) Tentukan sama ada setiap titik pusingan itu ialah titik maksimum atau titik minimum. Determine whether each of the turning points is a maximum point or a minimum point. [3 markah/marks] Strategi A+ Maths Tam Tg5-02_vim_3p(17-49).indd 48 26/10/2023 12:12:30 PM U BAKTI SDN. BHD. PENERBIT ILMU BAKTI SMU BAKTI SDN. BHD. PENERBIT ILMU BAKTIILMU BAKTI SDN. BHD. PENERBIT ILMU BAKIT ILMU BAKTI SDN. BHD. PENERBIT ILMU BA